Chapter 12 Introduction To Three Dimensional Geometry

Given:

A rectangular prism is placed in a three-dimensional coordinate system with one vertex at the origin O, as shown in the figure.

The coordinates of the vertex P are (2, 4, 5).

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To Find:

The coordinates of the vertex F.

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Solution:

From the figure, we can analyze the position of the vertices of the rectangular prism. The point P has coordinates (x, y, z), which represent the perpendicular distances of P from the YZ-plane, XZ-plane, and XY-plane, respectively.

We are given that the coordinates of P are (2, 4, 5). This means:

• The x-coordinate is 2.
• The y-coordinate is 4.
• The z-coordinate is 5.

The vertex F lies in the XZ-plane. A key property of any point in the XZ-plane is that its y-coordinate is always 0.

The vertex F is part of the same rectangular prism as P. As shown in the figure, F is the point obtained by moving from P parallel to the Y-axis until we reach the XZ-plane. This movement only changes the y-coordinate, while the x and z coordinates remain the same.

Therefore, if the coordinates of P are (x, y, z), the coordinates of F must be (x, 0, z).

Substituting the given coordinates of P = (2, 4, 5):

The x-coordinate of F is the same as P's, which is 2.

The y-coordinate of F is 0 (since it is in the XZ-plane).

The z-coordinate of F is the same as P's, which is 5.

So, the coordinates of F are (2, 0, 5).

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Final Answer:

The coordinates of F are (2, 0, 5).

Given:

Two points in a three-dimensional space: P(–3, 1, 2) and Q(–3, 1, – 2).

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To Find:

The octant in which each of the given points lies.

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Solution:

The octant of a point is determined by the signs of its x, y, and z coordinates. The sign conventions for the eight octants are summarized in the table below:

Octant	x-coordinate	y-coordinate	z-coordinate
I	+	+	+
II	–	+	+
III	–	–	+
IV	+	–	+
V	+	+	–
VI	–	+	–
VII	–	–	–
VIII	+	–	–

1. For the point (–3, 1, 2):

The x-coordinate is –3 (Negative).

The y-coordinate is 1 (Positive).

The z-coordinate is 2 (Positive).

The signs of the coordinates are (–, +, +). Referring to the table, this sign pattern corresponds to Octant II.

2. For the point (–3, 1, – 2):

The x-coordinate is –3 (Negative).

The y-coordinate is 1 (Positive).

The z-coordinate is –2 (Negative).

The signs of the coordinates are (–, +, –). Referring to the table, this sign pattern corresponds to Octant VI.

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Final Answer:

The point (–3, 1, 2) lies in Octant II.

The point (–3, 1, – 2) lies in Octant VI.

Given:

A point lies on the x-axis.

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To Find:

The y-coordinate and z-coordinate of the point.

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Solution:

In a three-dimensional Cartesian coordinate system, a point's location is described by an ordered triplet $(x, y, z)$.

The x-axis is the line where both the y-coordinate and the z-coordinate are zero. Any point on the x-axis can be represented in the form $(x, 0, 0)$, where $x$ is a real number representing the point's signed distance from the origin along the x-axis.

For a point to be on the x-axis, its perpendicular distance to the XZ-plane must be zero, which means its y-coordinate is 0. Similarly, its perpendicular distance to the XY-plane must be zero, which means its z-coordinate is 0.

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Final Answer:

If a point is on the x-axis, its y-coordinate is 0 and its z-coordinate is 0.

Given:

A point is in the XZ-plane.

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To Determine:

The value of its y-coordinate.

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Solution:

The XZ-plane is the plane that contains both the x-axis and the z-axis. By definition, every point lying in the XZ-plane has a perpendicular distance of zero from this plane.

The perpendicular distance of any point $(x, y, z)$ from the XZ-plane is given by the absolute value of its y-coordinate, $|y|$.

Since the point is in the XZ-plane, this distance must be zero.

$|y| = 0$

This implies that $y = 0$.

Therefore, any point in the XZ-plane can be represented in the form $(x, 0, z)$.

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Final Answer:

If a point is in the XZ-plane, its y-coordinate is 0.

Given:

A set of eight points in 3D space.

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To Determine:

The octant in which each point lies.

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Solution:

The octant of a point is determined by the signs of its x, y, and z coordinates. We can analyze each point based on the sign pattern (+ or –).

Point	Sign Pattern (x, y, z)	Octant
(1, 2, 3)	(+, +, +)	I
(4, –2, 3)	(+, –, +)	IV
(4, –2, –5)	(+, –, –)	VIII
(4, 2, –5)	(+, +, –)	V
(–4, 2, –5)	(–, +, –)	VI
(–4, 2, 5)	(–, +, +)	II
(–3, –1, 6)	(–, –, +)	III
(–2, –4, –7)	(–, –, –)	VII

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Final Answer:

• (1, 2, 3) lies in Octant I.
• (4, –2, 3) lies in Octant IV.
• (4, –2, –5) lies in Octant VIII.
• (4, 2, –5) lies in Octant V.
• (–4, 2, –5) lies in Octant VI.
• (–4, 2, 5) lies in Octant II.
• (–3, –1, 6) lies in Octant III.
• (–2, –4, –7) lies in Octant VII.

(i) The x-axis and y-axis taken together determine a plane known as the XY-plane.

This plane is defined by the equation $z=0$.

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(ii) The coordinates of points in the XY-plane are of the form (x, y, 0).

Since the XY-plane is defined by $z=0$, the z-coordinate of any point on this plane must be zero.

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(iii) Coordinate planes divide the space into eight octants.

The three mutually perpendicular coordinate planes (XY, YZ, and XZ) intersect at the origin and divide the three-dimensional space into eight distinct regions.

Given:

Two points in 3-dimensional space: P(1, –3, 4) and Q (– 4, 1, 2).

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To Find:

The distance between the points P and Q.

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Solution:

Let the coordinates of point P be $(x_1, y_1, z_1) = (1, -3, 4)$.

Let the coordinates of point Q be $(x_2, y_2, z_2) = (-4, 1, 2)$.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in 3-dimensional space is given by the distance formula:

$\displaystyle PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Substitute the given coordinates into the formula:

$\displaystyle PQ = \sqrt{(-4 - 1)^2 + (1 - (-3))^2 + (2 - 4)^2}$

Simplify the terms inside the square root:

$\displaystyle PQ = \sqrt{(-5)^2 + (1 + 3)^2 + (-2)^2}$

$\displaystyle PQ = \sqrt{(-5)^2 + (4)^2 + (-2)^2}$

Calculate the squares:

$(-5)^2 = 25$

$4^2 = 16$

$(-2)^2 = 4$

Substitute these values back into the formula:

$\displaystyle PQ = \sqrt{25 + 16 + 4}$

Add the numbers inside the square root:

$25 + 16 + 4 = 41 + 4 = 45$

$\displaystyle PQ = \sqrt{45}$

Simplify the square root of 45: $\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}$.

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Final Answer:

The distance between points P and Q is $3\sqrt{5}$ units.

Given:

Three points in 3-dimensional space: P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1).

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To Show:

The points P, Q, and R are collinear.

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Solution:

Three points P, Q, and R are collinear if the sum of the distances between two pairs of points is equal to the distance between the remaining pair of points. That is, $PQ + QR = PR$ or $PQ + PR = QR$ or $QR + PR = PQ$.

Let's calculate the distances between each pair of points using the distance formula:

$\displaystyle \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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Distance PQ:

P$(x_1, y_1, z_1) = (-2, 3, 5)$

Q$(x_2, y_2, z_2) = (1, 2, 3)$

$\displaystyle PQ = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2}$

$\displaystyle PQ = \sqrt{(1 + 2)^2 + (-1)^2 + (-2)^2}$

$\displaystyle PQ = \sqrt{3^2 + (-1)^2 + (-2)^2}$

$\displaystyle PQ = \sqrt{9 + 1 + 4} = \sqrt{14}$

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Distance QR:

Q$(x_1, y_1, z_1) = (1, 2, 3)$

R$(x_2, y_2, z_2) = (7, 0, -1)$

$\displaystyle QR = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2}$

$\displaystyle QR = \sqrt{6^2 + (-2)^2 + (-4)^2}$

$\displaystyle QR = \sqrt{36 + 4 + 16} = \sqrt{56}$

Simplify the square root: $\sqrt{56} = \sqrt{4 \times 14} = \sqrt{4} \times \sqrt{14} = 2\sqrt{14}$.

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Distance PR:

P$(x_1, y_1, z_1) = (-2, 3, 5)$

R$(x_2, y_2, z_2) = (7, 0, -1)$

$\displaystyle PR = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2}$

$\displaystyle PR = \sqrt{(7 + 2)^2 + (-3)^2 + (-6)^2}$

$\displaystyle PR = \sqrt{9^2 + (-3)^2 + (-6)^2}$

$\displaystyle PR = \sqrt{81 + 9 + 36} = \sqrt{126}$

Simplify the square root: $\sqrt{126} = \sqrt{9 \times 14} = \sqrt{9} \times \sqrt{14} = 3\sqrt{14}$.

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Now, check if the sum of two distances equals the third distance.

Consider $PQ + QR = \sqrt{14} + 2\sqrt{14} = (1 + 2)\sqrt{14} = 3\sqrt{14}$.

Comparing this with PR from equation (3), we see that $PQ + QR = PR$.

Since the sum of the distances between P and Q, and Q and R is equal to the distance between P and R, the points P, Q, and R are collinear. Furthermore, since PQ + QR = PR, the point Q lies between P and R.

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Conclusion:

Since $PQ + QR = PR$, the points P, Q, and R are collinear.

Given:

The coordinates of three points are A(3, 6, 9), B(10, 20, 30), and C(25, –41, 5).

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To Find:

Whether the given points are the vertices of a right-angled triangle.

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Solution:

To determine if the triangle formed by the points A, B, and C is a right-angled triangle, we can use the Pythagorean theorem. A triangle is right-angled if the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides.

First, we need to find the square of the lengths of the three sides of the triangle ABC using the distance formula in three dimensions.

The square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$

1. Square of the length of side AB:

$AB^2 = (10 - 3)^2 + (20 - 6)^2 + (30 - 9)^2$

$AB^2 = (7)^2 + (14)^2 + (21)^2$

$AB^2 = 49 + 196 + 441$

$AB^2 = 686$

2. Square of the length of side BC:

$BC^2 = (25 - 10)^2 + (-41 - 20)^2 + (5 - 30)^2$

$BC^2 = (15)^2 + (-61)^2 + (-25)^2$

$BC^2 = 225 + 3721 + 625$

$BC^2 = 4571$

3. Square of the length of side AC:

$AC^2 = (25 - 3)^2 + (-41 - 6)^2 + (5 - 9)^2$

$AC^2 = (22)^2 + (-47)^2 + (-4)^2$

$AC^2 = 484 + 2209 + 16$

$AC^2 = 2709$

Now, we check if the Pythagorean theorem holds true. The theorem states that for a right-angled triangle, the sum of the squares of two sides equals the square of the third (longest) side.

Let's check the sum of the squares of the two shorter sides, AB and AC:

$AB^2 + AC^2 = 686 + 2709 = 3395$

We compare this sum with the square of the longest side, BC:

$BC^2 = 4571$

Clearly, $AB^2 + AC^2 \neq BC^2$, since $3395 \neq 4571$.

Since the sum of the squares of two sides is not equal to the square of the third side, the triangle ABC does not satisfy the Pythagorean theorem.

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Final Answer:

The points A (3, 6, 9), B (10, 20, 30), and C (25, –41, 5) are not the vertices of a right-angled triangle.

Given:

A set of points P, where each point P satisfies the condition $PA^2 + PB^2 = 2k^2$.

The coordinates of point A are (3, 4, 5).

The coordinates of point B are (–1, 3, –7).

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To Find:

The equation of the set of points P.

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Solution:

Let the coordinates of any point P in the set be (x, y, z).

We are given the fixed points A(3, 4, 5) and B(–1, 3, –7).

We use the distance formula in three dimensions. The square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$.

First, we find an expression for $PA^2$:

$PA^2 = (x - 3)^2 + (y - 4)^2 + (z - 5)^2$

Expanding the terms, we get:

$PA^2 = (x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 - 10z + 25)$

$PA^2 = x^2 + y^2 + z^2 - 6x - 8y - 10z + 50$

Next, we find an expression for $PB^2$:

$PB^2 = (x - (-1))^2 + (y - 3)^2 + (z - (-7))^2$

$PB^2 = (x + 1)^2 + (y - 3)^2 + (z + 7)^2$

Expanding the terms, we get:

$PB^2 = (x^2 + 2x + 1) + (y^2 - 6y + 9) + (z^2 + 14z + 49)$

$PB^2 = x^2 + y^2 + z^2 + 2x - 6y + 14z + 59$

Now, we use the given condition:

$PA^2 + PB^2 = 2k^2$

Substitute the expanded expressions for $PA^2$ and $PB^2$ into this equation:

$(x^2 + y^2 + z^2 - 6x - 8y - 10z + 50) + (x^2 + y^2 + z^2 + 2x - 6y \ $$ + 14z + 59) = 2k^2$

Now, we combine the like terms:

$(x^2 + x^2) + (y^2 + y^2) + (z^2 + z^2) + (-6x + 2x) + (-8y - 6y) + (-10z \ $$ + 14z) + (50 + 59) = 2k^2$

$2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = 2k^2$

This is the required equation that represents the set of points P. It is the equation of a sphere.

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Final Answer:

The equation of the set of points P is $2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = 2k^2$.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three dimensions is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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(i) Points are $(2, 3, 5)$ and $(4, 3, 1)$.

Let $(x_1, y_1, z_1) = (2, 3, 5)$ and $(x_2, y_2, z_2) = (4, 3, 1)$.

Distance $d = \sqrt{(4 - 2)^2 + (3 - 3)^2 + (1 - 5)^2}$

$d = \sqrt{(2)^2 + (0)^2 + (-4)^2}$

$d = \sqrt{4 + 0 + 16}$

$d = \sqrt{20}$

$d = \sqrt{4 \times 5}$

$d = 2\sqrt{5}$

The distance between $(2, 3, 5)$ and $(4, 3, 1)$ is $2\sqrt{5}$.

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(ii) Points are $(-3, 7, 2)$ and $(2, 4, -1)$.

Let $(x_1, y_1, z_1) = (-3, 7, 2)$ and $(x_2, y_2, z_2) = (2, 4, -1)$.

Distance $d = \sqrt{(2 - (-3))^2 + (4 - 7)^2 + (-1 - 2)^2}$

$d = \sqrt{(2 + 3)^2 + (-3)^2 + (-3)^2}$

$d = \sqrt{(5)^2 + 9 + 9}$

$d = \sqrt{25 + 9 + 9}$

$d = \sqrt{43}$

The distance between $(-3, 7, 2)$ and $(2, 4, -1)$ is $\sqrt{43}$.

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(iii) Points are $(-1, 3, -4)$ and $(1, -3, 4)$.

Let $(x_1, y_1, z_1) = (-1, 3, -4)$ and $(x_2, y_2, z_2) = (1, -3, 4)$.

Distance $d = \sqrt{(1 - (-1))^2 + (-3 - 3)^2 + (4 - (-4))^2}$

$d = \sqrt{(1 + 1)^2 + (-6)^2 + (4 + 4)^2}$

$d = \sqrt{(2)^2 + 36 + (8)^2}$

$d = \sqrt{4 + 36 + 64}$

$d = \sqrt{104}$

$d = \sqrt{4 \times 26}$

$d = 2\sqrt{26}$

The distance between $(-1, 3, -4)$ and $(1, -3, 4)$ is $2\sqrt{26}$.

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(iv) Points are $(2, -1, 3)$ and $(-2, 1, 3)$.

Let $(x_1, y_1, z_1) = (2, -1, 3)$ and $(x_2, y_2, z_2) = (-2, 1, 3)$.

Distance $d = \sqrt{(-2 - 2)^2 + (1 - (-1))^2 + (3 - 3)^2}$

$d = \sqrt{(-4)^2 + (1 + 1)^2 + (0)^2}$

$d = \sqrt{16 + (2)^2 + 0}$

$d = \sqrt{16 + 4}$

$d = \sqrt{20}$

$d = \sqrt{4 \times 5}$

$d = 2\sqrt{5}$

The distance between $(2, -1, 3)$ and $(-2, 1, 3)$ is $2\sqrt{5}$.

Let the given points be A($-2, 3, 5$), B($1, 2, 3$), and C($7, 0, -1$).

To show that these points are collinear, we need to verify if the sum of the distances between two pairs of points is equal to the distance between the third pair.

The distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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Calculate the distance AB:

A($-2, 3, 5$) and B($1, 2, 3$)

$AB = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2}$

$AB = \sqrt{(1 + 2)^2 + (-1)^2 + (-2)^2}$

$AB = \sqrt{(3)^2 + 1 + 4}$

$AB = \sqrt{9 + 1 + 4}$

$AB = \sqrt{14}$

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Calculate the distance BC:

B($1, 2, 3$) and C($7, 0, -1$)

$BC = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2}$

$BC = \sqrt{(6)^2 + (-2)^2 + (-4)^2}$

$BC = \sqrt{36 + 4 + 16}$

$BC = \sqrt{56}$

$BC = \sqrt{4 \times 14}$

$BC = 2\sqrt{14}$

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Calculate the distance AC:

A($-2, 3, 5$) and C($7, 0, -1$)

$AC = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2}$

$AC = \sqrt{(7 + 2)^2 + (-3)^2 + (-6)^2}$

$AC = \sqrt{(9)^2 + 9 + 36}$

$AC = \sqrt{81 + 9 + 36}$

$AC = \sqrt{126}$

$AC = \sqrt{9 \times 14}$

$AC = 3\sqrt{14}$

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Now, let's check if the sum of any two distances equals the third distance:

$AB + BC = \sqrt{14} + 2\sqrt{14} = (1 + 2)\sqrt{14} = 3\sqrt{14}$

We see that $AC = 3\sqrt{14}$.

Therefore, $AB + BC = AC$.

Since the sum of the distances between two pairs of points is equal to the distance between the third pair, the points A, B, and C are collinear.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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(i) Verify if the points (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.

Let the points be A(0, 7, -10), B(1, 6, -6), and C(4, 9, -6).

Calculate the length of side AB:

$AB = \sqrt{(1 - 0)^2 + (6 - 7)^2 + (-6 - (-10))^2}$

$AB = \sqrt{(1)^2 + (-1)^2 + (-6 + 10)^2}$

$AB = \sqrt{1 + 1 + (4)^2}$

$AB = \sqrt{1 + 1 + 16}$

$AB = \sqrt{18}$

Calculate the length of side BC:

$BC = \sqrt{(4 - 1)^2 + (9 - 6)^2 + (-6 - (-6))^2}$

$BC = \sqrt{(3)^2 + (3)^2 + (-6 + 6)^2}$

$BC = \sqrt{9 + 9 + (0)^2}$

$BC = \sqrt{9 + 9 + 0}$

$BC = \sqrt{18}$

Calculate the length of side AC:

$AC = \sqrt{(4 - 0)^2 + (9 - 7)^2 + (-6 - (-10))^2}$

$AC = \sqrt{(4)^2 + (2)^2 + (-6 + 10)^2}$

$AC = \sqrt{16 + 4 + (4)^2}$

$AC = \sqrt{16 + 4 + 16}$

$AC = \sqrt{36}$

$AC = 6$

Since $AB = \sqrt{18}$ and $BC = \sqrt{18}$, we have $AB = BC$.

As two sides of the triangle ABC are equal in length, the triangle ABC is an isosceles triangle.

Hence, the verification is complete.

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(ii) Verify if the points (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.

Let the points be P(0, 7, 10), Q(-1, 6, 6), and R(-4, 9, 6).

To check if it's a right-angled triangle, we can use the Pythagorean theorem: the square of the longest side should be equal to the sum of the squares of the other two sides.

Calculate the square of the length of side PQ:

$PQ^2 = (-1 - 0)^2 + (6 - 7)^2 + (6 - 10)^2$

$PQ^2 = (-1)^2 + (-1)^2 + (-4)^2$

$PQ^2 = 1 + 1 + 16$

$PQ^2 = 18$

Calculate the square of the length of side QR:

$QR^2 = (-4 - (-1))^2 + (9 - 6)^2 + (6 - 6)^2$

$QR^2 = (-4 + 1)^2 + (3)^2 + (0)^2$

$QR^2 = (-3)^2 + 9 + 0$

$QR^2 = 9 + 9 + 0$

$QR^2 = 18$

Calculate the square of the length of side PR:

$PR^2 = (-4 - 0)^2 + (9 - 7)^2 + (6 - 10)^2$

$PR^2 = (-4)^2 + (2)^2 + (-4)^2$

$PR^2 = 16 + 4 + 16$

$PR^2 = 36$

Now, let's check the sum of the squares of the two shorter sides:

$PQ^2 + QR^2 = 18 + 18 = 36$

We see that $PQ^2 + QR^2 = 36$, which is equal to $PR^2$.

Since the sum of the squares of two sides is equal to the square of the third side ($PQ^2 + QR^2 = PR^2$), the triangle PQR is a right angled triangle by the converse of the Pythagorean theorem. The right angle is at vertex Q.

Hence, the verification is complete.

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(iii) Verify if the points (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Let the points be A(-1, 2, 1), B(1, -2, 5), C(4, -7, 8), and D(2, -3, 4).

To show that these points are the vertices of a parallelogram, we can verify if the opposite sides have equal length.

Calculate the length of side AB:

$AB = \sqrt{(1 - (-1))^2 + (-2 - 2)^2 + (5 - 1)^2}$

$AB = \sqrt{(1 + 1)^2 + (-4)^2 + (4)^2}$

$AB = \sqrt{(2)^2 + 16 + 16}$

$AB = \sqrt{4 + 16 + 16}$

$AB = \sqrt{36}$

$AB = 6$

Calculate the length of side BC:

$BC = \sqrt{(4 - 1)^2 + (-7 - (-2))^2 + (8 - 5)^2}$

$BC = \sqrt{(3)^2 + (-7 + 2)^2 + (3)^2}$

$BC = \sqrt{9 + (-5)^2 + 9}$

$BC = \sqrt{9 + 25 + 9}$

$BC = \sqrt{43}$

Calculate the length of side CD:

$CD = \sqrt{(2 - 4)^2 + (-3 - (-7))^2 + (4 - 8)^2}$

$CD = \sqrt{(-2)^2 + (-3 + 7)^2 + (-4)^2}$

$CD = \sqrt{4 + (4)^2 + 16}$

$CD = \sqrt{4 + 16 + 16}$

$CD = \sqrt{36}$

$CD = 6$

Calculate the length of side DA:

$DA = \sqrt{(-1 - 2)^2 + (2 - (-3))^2 + (1 - 4)^2}$

$DA = \sqrt{(-3)^2 + (2 + 3)^2 + (-3)^2}$

$DA = \sqrt{9 + (5)^2 + 9}$

$DA = \sqrt{9 + 25 + 9}$

$DA = \sqrt{43}$

We observe that $AB = 6$ and $CD = 6$, so $AB = CD$.

We also observe that $BC = \sqrt{43}$ and $DA = \sqrt{43}$, so $BC = DA$.

Since both pairs of opposite sides are equal in length, the quadrilateral ABCD is a parallelogram.

Hence, the verification is complete.

Let the set of points be denoted by P(x, y, z). Let the given points be A(1, 2, 3) and B(3, 2, -1).

According to the problem, the distance from P to A is equal to the distance from P to B.

$PA = PB$

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, which is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$, we can write:

Distance PA = $\sqrt{(x - 1)^2 + (y - 2)^2 + (z - 3)^2}$

Distance PB = $\sqrt{(x - 3)^2 + (y - 2)^2 + (z - (-1))^2} \ $$ = \sqrt{(x - 3)^2 + (y - 2)^2 + (z + 1)^2}$

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Since PA = PB, we have:

$\sqrt{(x - 1)^2 + (y - 2)^2 + (z - 3)^2} = \sqrt{(x - 3)^2 + (y - 2)^2 + (z + 1)^2}$

Squaring both sides to remove the square root:

$(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 3)^2 + (y - 2)^2 + (z + 1)^2$

Expand the squared terms:

$(x^2 - 2x + 1) + (y^2 - 4y + 4) + (z^2 - 6z + 9) = (x^2 - 6x + 9) \ $$ + (y^2 - 4y + 4) + (z^2 + 2z + 1)$

Cancel the common terms $x^2, y^2, z^2, -4y, 4$ from both sides:

$-2x + 1 + z^2 - 6z + 9 = -6x + 9 + z^2 + 2z + 1$

Cancel $z^2$ and $1+9=10$ from both sides:

$-2x - 6z = -6x + 2z$

Move all terms involving x and z to one side:

$-2x + 6x - 6z - 2z = 0$

$4x - 8z = 0$

Divide the entire equation by 4:

$\frac{4x}{4} - \frac{8z}{4} = \frac{0}{4}$

$x - 2z = 0$

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The equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -1) is $x - 2z = 0$.

Let P(x, y, z) be any point in the set of points. Let A be the point (4, 0, 0) and B be the point (-4, 0, 0).

According to the problem, the sum of the distances from P to A and from P to B is equal to 10.

$PA + PB = 10$

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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Using the distance formula, we can write PA and PB as:

$PA = \sqrt{(x - 4)^2 + (y - 0)^2 + (z - 0)^2} = \sqrt{(x - 4)^2 + y^2 + z^2}$

$PB = \sqrt{(x - (-4))^2 + (y - 0)^2 + (z - 0)^2} = \sqrt{(x + 4)^2 + y^2 + z^2}$

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Substituting these into the given condition:

$\sqrt{(x - 4)^2 + y^2 + z^2} + \sqrt{(x + 4)^2 + y^2 + z^2} = 10$

Isolate one square root term:

$\sqrt{(x - 4)^2 + y^2 + z^2} = 10 - \sqrt{(x + 4)^2 + y^2 + z^2}$

Square both sides:

$(x - 4)^2 + y^2 + z^2 = \left(10 - \sqrt{(x + 4)^2 + y^2 + z^2}\right)^2$

Expand both sides:

$(x^2 - 8x + 16) + y^2 + z^2 = 100 - 20\sqrt{(x + 4)^2 + y^2 + z^2} \ $$ + ((x + 4)^2 + y^2 + z^2)$

$x^2 - 8x + 16 + y^2 + z^2 = 100 - 20\sqrt{(x + 4)^2 + y^2 + z^2} \ $$ + (x^2 + 8x + 16) + y^2 + z^2$

Cancel $x^2, 16, y^2, z^2$ from both sides:

$-8x = 100 - 20\sqrt{(x + 4)^2 + y^2 + z^2} + 8x$

Rearrange to isolate the remaining square root term:

$20\sqrt{(x + 4)^2 + y^2 + z^2} = 100 + 8x + 8x$

$20\sqrt{(x + 4)^2 + y^2 + z^2} = 100 + 16x$

Divide the equation by 4:

$5\sqrt{(x + 4)^2 + y^2 + z^2} = 25 + 4x$

Square both sides again:

$(5\sqrt{(x + 4)^2 + y^2 + z^2})^2 = (25 + 4x)^2$

$25 \left((x + 4)^2 + y^2 + z^2\right) = (25)^2 + 2(25)(4x) + (4x)^2$

$25 (x^2 + 8x + 16 + y^2 + z^2) = 625 + 200x + 16x^2$

Distribute 25 on the left side:

$25x^2 + 200x + 400 + 25y^2 + 25z^2 = 625 + 200x + 16x^2$

Cancel $200x$ from both sides:

$25x^2 + 400 + 25y^2 + 25z^2 = 625 + 16x^2$

Rearrange the terms to group x, y, z terms on one side and the constant on the other:

$25x^2 - 16x^2 + 25y^2 + 25z^2 = 625 - 400$

Simplify:

$9x^2 + 25y^2 + 25z^2 = 225$

Divide the entire equation by 225 to put it in standard form:

$\frac{9x^2}{225} + \frac{25y^2}{225} + \frac{25z^2}{225} = \frac{225}{225}$

Simplify the fractions:

$\frac{x^2}{25} + \frac{y^2}{9} + \frac{z^2}{9} = 1$

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The equation of the set of points P such that the sum of the distances from A(4, 0, 0) and B(-4, 0, 0) is equal to 10 is $\frac{x^2}{25} + \frac{y^2}{9} + \frac{z^2}{9} = 1$. This equation represents an ellipsoid.

Let the given points be A$(1, -2, 3)$ and B$(3, 4, -5)$. We want to find the coordinates of the point P that divides the line segment AB in the ratio 2 : 3.

Let $(x_1, y_1, z_1) = (1, -2, 3)$ and $(x_2, y_2, z_2) = (3, 4, -5)$. The ratio is $m : n = 2 : 3$.

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(i) Internal division:

The coordinates of a point P that divides the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ internally in the ratio $m : n$ are given by the section formula:

$P = \left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}, \frac{mz_2 + nz_1}{m + n}\right)$

Here, $m = 2$, $n = 3$, $(x_1, y_1, z_1) = (1, -2, 3)$, and $(x_2, y_2, z_2) = (3, 4, -5)$.

x-coordinate of P = $\frac{(2)(3) + (3)(1)}{2 + 3} = \frac{6 + 3}{5} = \frac{9}{5}$

y-coordinate of P = $\frac{(2)(4) + (3)(-2)}{2 + 3} = \frac{8 - 6}{5} = \frac{2}{5}$

z-coordinate of P = $\frac{(2)(-5) + (3)(3)}{2 + 3} = \frac{-10 + 9}{5} = \frac{-1}{5}$

So, the coordinates of the point P that divides the line segment internally in the ratio 2 : 3 are $\left(\frac{9}{5}, \frac{2}{5}, -\frac{1}{5}\right)$.

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(ii) External division:

The coordinates of a point P that divides the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ externally in the ratio $m : n$ are given by the section formula:

$P = \left(\frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n}, \frac{mz_2 - nz_1}{m - n}\right)$

Here, $m = 2$, $n = 3$, $(x_1, y_1, z_1) = (1, -2, 3)$, and $(x_2, y_2, z_2) = (3, 4, -5)$.

x-coordinate of P = $\frac{(2)(3) - (3)(1)}{2 - 3} = \frac{6 - 3}{-1} = \frac{3}{-1} = -3$

y-coordinate of P = $\frac{(2)(4) - (3)(-2)}{2 - 3} = \frac{8 - (-6)}{-1} = \frac{8 + 6}{-1} = \frac{14}{-1} = -14$

z-coordinate of P = $\frac{(2)(-5) - (3)(3)}{2 - 3} = \frac{-10 - 9}{-1} = \frac{-19}{-1} = 19$

So, the coordinates of the point P that divides the line segment externally in the ratio 2 : 3 are $(-3, -14, 19)$.

Let the given points be A$(-4, 6, 10)$, B$(2, 4, 6)$, and C$(14, 0, -2)$.

To prove that these points are collinear using the section formula, we need to show that one of the points (say B) divides the line segment joining the other two points (A and C) in a specific ratio.

Let's assume that the point B$(2, 4, 6)$ divides the line segment joining A$(-4, 6, 10)$ and C$(14, 0, -2)$ in the ratio $m : n$.

The coordinates of a point dividing the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $m : n$ are given by the section formula:

$\left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}, \frac{mz_2 + nz_1}{m + n}\right)$

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Using the coordinates of B as $(x, y, z) = (2, 4, 6)$, and A as $(x_1, y_1, z_1) = (-4, 6, 10)$, and C as $(x_2, y_2, z_2) = (14, 0, -2)$, we set up the equations:

For the x-coordinate:

$2 = \frac{m(14) + n(-4)}{m + n}$

$2(m + n) = 14m - 4n$

$2m + 2n = 14m - 4n$

$2n + 4n = 14m - 2m$

$6n = 12m$

For the y-coordinate:

$4 = \frac{m(0) + n(6)}{m + n}$

$4(m + n) = 6n$

$4m + 4n = 6n$

$4m = 6n - 4n$

$4m = 2n$

For the z-coordinate:

$6 = \frac{m(-2) + n(10)}{m + n}$

$6(m + n) = -2m + 10n$

$6m + 6n = -2m + 10n$

$6m + 2m = 10n - 6n$

$8m = 4n$

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From equations (i), (ii), and (iii), we see that the ratio $m : n$ is $1 : 2$ for all three coordinates.

This means that point B$(2, 4, 6)$ divides the line segment joining A$(-4, 6, 10)$ and C$(14, 0, -2)$ internally in the ratio $1 : 2$.

Since B lies on the line segment AC, the points A, B, and C are collinear.

Hence, the statement is proved using the section formula.

Let the vertices of the triangle be A$(x_1, y_1, z_1)$, B$(x_2, y_2, z_2)$, and C$(x_3, y_3, z_3)$.

The centroid of a triangle is the point of intersection of its medians. A median joins a vertex to the midpoint of the opposite side.

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Let D be the midpoint of the side BC. The coordinates of the midpoint of a line segment joining $(x_a, y_a, z_a)$ and $(x_b, y_b, z_b)$ are given by $\left(\frac{x_a + x_b}{2}, \frac{y_a + y_b}{2}, \frac{z_a + z_b}{2}\right)$.

Using this, the coordinates of D are:

$D = \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2}\right)$

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The centroid G divides the median AD in the ratio $2 : 1$. We can use the section formula to find the coordinates of G.

The section formula for a point dividing the line segment joining $(X_1, Y_1, Z_1)$ and $(X_2, Y_2, Z_2)$ in the ratio $m : n$ internally is $\left(\frac{mX_2 + nX_1}{m + n}, \frac{mY_2 + nY_1}{m + n}, \frac{mZ_2 + nZ_1}{m + n}\right)$.

Here, $(X_1, Y_1, Z_1) = A(x_1, y_1, z_1)$, $(X_2, Y_2, Z_2) = D\left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2}\right)$, and the ratio is $m:n = 2:1$.

The coordinates of the centroid G are:

x-coordinate of G = $\frac{2 \left(\frac{x_2 + x_3}{2}\right) + 1 (x_1)}{2 + 1} = \frac{x_2 + x_3 + x_1}{3} = \frac{x_1 + x_2 + x_3}{3}$

y-coordinate of G = $\frac{2 \left(\frac{y_2 + y_3}{2}\right) + 1 (y_1)}{2 + 1} = \frac{y_2 + y_3 + y_1}{3} = \frac{y_1 + y_2 + y_3}{3}$

z-coordinate of G = $\frac{2 \left(\frac{z_2 + z_3}{2}\right) + 1 (z_1)}{2 + 1} = \frac{z_2 + z_3 + z_1}{3} = \frac{z_1 + z_2 + z_3}{3}$

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Thus, the coordinates of the centroid of the triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ are $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)$.

Let the given points be A$(4, 8, 10)$ and B$(6, 10, -8)$.

We want to find the ratio in which the line segment AB is divided by the YZ-plane.

Let the point of division be P. Any point on the YZ-plane has its x-coordinate equal to 0. So, the coordinates of P are $(0, y, z)$.

Let the ratio in which P divides the line segment AB be $k : 1$.

Using the section formula, the coordinates of P are given by:

$P = \left(\frac{k x_2 + 1 x_1}{k + 1}, \frac{k y_2 + 1 y_1}{k + 1}, \frac{k z_2 + 1 z_1}{k + 1}\right)$

Here, $(x_1, y_1, z_1) = (4, 8, 10)$ and $(x_2, y_2, z_2) = (6, 10, -8)$.

So, the coordinates of P are:

$P = \left(\frac{k(6) + 1(4)}{k + 1}, \frac{k(10) + 1(8)}{k + 1}, \frac{k(-8) + 1(10)}{k + 1}\right)$

$P = \left(\frac{6k + 4}{k + 1}, \frac{10k + 8}{k + 1}, \frac{-8k + 10}{k + 1}\right)$

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Since the point P lies on the YZ-plane, its x-coordinate must be 0.

Therefore, we have:

$\frac{6k + 4}{k + 1} = 0$

For this fraction to be zero, the numerator must be zero (provided the denominator is not zero):

$6k + 4 = 0$

$6k = -4$

$k = \frac{-4}{6}$

$k = -\frac{2}{3}$

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The ratio is $k : 1 = -\frac{2}{3} : 1$. Multiplying both parts by 3, we get the ratio $-2 : 3$.

A negative ratio indicates external division.

Thus, the YZ-plane divides the line segment joining the points (4, 8, 10) and (6, 10, -8) in the ratio 2 : 3 externally.

Let the given points be A$(-2, 3, 5)$ and B$(1, -4, 6)$. We want to find the coordinates of the point P that divides the line segment AB in the ratio 2 : 3.

Let $(x_1, y_1, z_1) = (-2, 3, 5)$ and $(x_2, y_2, z_2) = (1, -4, 6)$. The ratio is $m : n = 2 : 3$.

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(i) Internal division:

The coordinates of a point P that divides the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ internally in the ratio $m : n$ are given by the section formula:

$P = \left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}, \frac{mz_2 + nz_1}{m + n}\right)$

Substitute the given values $m=2$, $n=3$, $(x_1, y_1, z_1) = (-2, 3, 5)$, and $(x_2, y_2, z_2) = (1, -4, 6)$ into the formula:

x-coordinate of P = $\frac{(2)(1) + (3)(-2)}{2 + 3} = \frac{2 - 6}{5} = \frac{-4}{5}$

y-coordinate of P = $\frac{(2)(-4) + (3)(3)}{2 + 3} = \frac{-8 + 9}{5} = \frac{1}{5}$

z-coordinate of P = $\frac{(2)(6) + (3)(5)}{2 + 3} = \frac{12 + 15}{5} = \frac{27}{5}$

So, the coordinates of the point P that divides the line segment internally in the ratio 2 : 3 are $\left(-\frac{4}{5}, \frac{1}{5}, \frac{27}{5}\right)$.

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(ii) External division:

The coordinates of a point P that divides the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ externally in the ratio $m : n$ are given by the section formula:

$P = \left(\frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n}, \frac{mz_2 - nz_1}{m - n}\right)$

Substitute the given values $m=2$, $n=3$, $(x_1, y_1, z_1) = (-2, 3, 5)$, and $(x_2, y_2, z_2) = (1, -4, 6)$ into the formula:

x-coordinate of P = $\frac{(2)(1) - (3)(-2)}{2 - 3} = \frac{2 - (-6)}{-1} = \frac{2 + 6}{-1} = \frac{8}{-1} = -8$

y-coordinate of P = $\frac{(2)(-4) - (3)(3)}{2 - 3} = \frac{-8 - 9}{-1} = \frac{-17}{-1} = 17$

z-coordinate of P = $\frac{(2)(6) - (3)(5)}{2 - 3} = \frac{12 - 15}{-1} = \frac{-3}{-1} = 3$

So, the coordinates of the point P that divides the line segment externally in the ratio 2 : 3 are $(-8, 17, 3)$.

Given:

Points P(3, 2, -4), Q(5, 4, -6), and R(9, 8, -10) are collinear.

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To Find:

The ratio in which Q divides the line segment PR.

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Solution:

Let the point Q(5, 4, -6) divide the line segment joining P(3, 2, -4) and R(9, 8, -10) in the ratio $k : 1$.

Using the section formula for internal division, the coordinates of the point dividing the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $k : 1$ are given by:

$\left(\frac{k x_2 + 1 x_1}{k + 1}, \frac{k y_2 + 1 y_1}{k + 1}, \frac{k z_2 + 1 z_1}{k + 1}\right)$

Here, $(x_1, y_1, z_1) = (3, 2, -4)$, $(x_2, y_2, z_2) = (9, 8, -10)$, and the point of division is Q(5, 4, -6).

Equating the coordinates of Q with the section formula:

For the x-coordinate:

For the y-coordinate:

For the z-coordinate:

Let's solve equation (i) for $k$:

$5(k + 1) = 9k + 3$

$5k + 5 = 9k + 3$

$5 - 3 = 9k - 5k$

$2 = 4k$

$k = \frac{2}{4} = \frac{1}{2}$

Let's verify this value of $k$ with equation (ii):

$4 = \frac{\frac{1}{2}(8) + 1(2)}{\frac{1}{2} + 1} = \frac{4 + 2}{\frac{3}{2}} = \frac{6}{\frac{3}{2}} = 6 \times \frac{2}{3} = 4$

The value matches.

Let's verify this value of $k$ with equation (iii):

$-6 = \frac{\frac{1}{2}(-10) + 1(-4)}{\frac{1}{2} + 1} = \frac{-5 - 4}{\frac{3}{2}} = \frac{-9}{\frac{3}{2}} = -9 \times \frac{2}{3} = -6$

The value matches.

Since $k = \frac{1}{2}$, the ratio $k : 1$ is $\frac{1}{2} : 1$, which is equivalent to $1 : 2$.

The positive value of $k$ indicates that Q divides PR internally.

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Thus, Q divides the line segment PR in the ratio 1 : 2 internally.

Given:

The two points forming the line segment are A$(-2, 4, 7)$ and B$(3, -5, 8)$.

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To Find:

The ratio in which the YZ-plane divides the line segment AB.

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Solution:

Let the point where the line segment AB is divided by the YZ-plane be P.

Any point on the YZ-plane has its x-coordinate equal to 0.

So, let the coordinates of the point P be $(0, y, z)$.

Let P divide the line segment joining A$(-2, 4, 7)$ and B$(3, -5, 8)$ in the ratio $k : 1$.

Using the section formula, the coordinates of the point P that divides the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $k : 1$ are given by:

$P = \left(\frac{k x_2 + 1 x_1}{k + 1}, \frac{k y_2 + 1 y_1}{k + 1}, \frac{k z_2 + 1 z_1}{k + 1}\right)$

Here, $(x_1, y_1, z_1) = (-2, 4, 7)$, $(x_2, y_2, z_2) = (3, -5, 8)$, and the point P is $(0, y, z)$.

Equating the coordinates of P with the section formula coordinates:

For the x-coordinate:

For the y-coordinate:

$y = \frac{k(-5) + 1(4)}{k + 1} = \frac{-5k + 4}{k + 1}$

For the z-coordinate:

$z = \frac{k(8) + 1(7)}{k + 1} = \frac{8k + 7}{k + 1}$

Now, we solve equation (i) for $k$:

$0 = \frac{3k - 2}{k + 1}$

This implies the numerator must be zero (assuming $k+1 \neq 0$):

$3k - 2 = 0$

$3k = 2$

$k = \frac{2}{3}$

The ratio in which P divides AB is $k : 1 = \frac{2}{3} : 1$.

Multiplying both parts of the ratio by 3, we get the ratio $2 : 3$.

Since the value of $k$ is positive, the division is internal.

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The YZ-plane divides the line segment formed by joining the points $(-2, 4, 7)$ and $(3, -5, 8)$ in the ratio 2 : 3 internally.

We can also find the coordinates of the point of division by substituting $k = \frac{2}{3}$ into the expressions for $y$ and $z$:

$y = \frac{-5(\frac{2}{3}) + 4}{\frac{2}{3} + 1} = \frac{-\frac{10}{3} + \frac{12}{3}}{\frac{5}{3}} = \frac{\frac{2}{3}}{\frac{5}{3}} = \frac{2}{5}$

$z = \frac{8(\frac{2}{3}) + 7}{\frac{2}{3} + 1} = \frac{\frac{16}{3} + \frac{21}{3}}{\frac{5}{3}} = \frac{\frac{37}{3}}{\frac{5}{3}} = \frac{37}{5}$

The point of division is $\left(0, \frac{2}{5}, \frac{37}{5}\right)$.

Given:

The three points are A$(2, -3, 4)$, B$(-1, 2, 1)$, and C$\left(0, \frac{1}{3}, 2\right)$.

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To Show:

The points A, B, and C are collinear using the section formula.

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Proof:

To show that the points A, B, and C are collinear using the section formula, we need to demonstrate that one of the points divides the line segment joining the other two points in a constant ratio.

Let's assume that the point B$(-1, 2, 1)$ divides the line segment joining A$(2, -3, 4)$ and C$\left(0, \frac{1}{3}, 2\right)$ in the ratio $k : 1$.

The section formula for the coordinates of a point dividing the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $k : 1$ is:

$\left(\frac{k x_2 + 1 x_1}{k + 1}, \frac{k y_2 + 1 y_1}{k + 1}, \frac{k z_2 + 1 z_1}{k + 1}\right)$

Here, $(x_1, y_1, z_1) = A(2, -3, 4)$, $(x_2, y_2, z_2) = C\left(0, \frac{1}{3}, 2\right)$, and the point of division is B$(-1, 2, 1)$.

Equating the coordinates of B with the section formula coordinates:

For the x-coordinate:

From (i):

$-1(k + 1) = 2$

$-k - 1 = 2$

$-k = 3$

$k = -3$

For the y-coordinate:

From (ii):

$2(k + 1) = \frac{k}{3} - 3$

$2k + 2 = \frac{k}{3} - 3$

Multiply by 3:

$3(2k + 2) = 3\left(\frac{k}{3} - 3\right)$

$6k + 6 = k - 9$

$6k - k = -9 - 6$

$5k = -15$

$k = \frac{-15}{5} = -3$

For the z-coordinate:

From (iii):

$1(k + 1) = 2k + 4$

$k + 1 = 2k + 4$

$1 - 4 = 2k - k$

$-3 = k$

$k = -3$

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Since the value of $k$ is the same ($k = -3$) when calculated from the x, y, and z coordinates, it means that point B divides the line segment joining A and C in the ratio $-3 : 1$. This indicates external division in the ratio 3 : 1.

Because B divides the line segment AC in a fixed ratio, the points A, B, and C lie on the same straight line.

Therefore, the points A $(2, -3, 4)$, B $(-1, 2, 1)$, and C$\left(0, \frac{1}{3}, 2\right)$ are collinear.

Hence, the statement is proved using the section formula.

Let the given points be P$(4, 2, -6)$ and Q$(10, -16, 6)$.

To trisect the line segment PQ means to divide it into three equal parts. This is achieved by two points that lie on the line segment.

Let these two points be A and B. If A and B trisect PQ, then A divides PQ in the ratio $1 : 2$ internally, and B divides PQ in the ratio $2 : 1$ internally. Alternatively, B is the midpoint of AQ.

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Finding the coordinates of the first point (A):

Point A divides the line segment PQ joining P$(4, 2, -6)$ and Q$(10, -16, 6)$ in the ratio $1 : 2$ internally.

Using the section formula for internal division for a point dividing the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $m : n$:

$\left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}, \frac{mz_2 + nz_1}{m + n}\right)$

Here, $(x_1, y_1, z_1) = (4, 2, -6)$, $(x_2, y_2, z_2) = (10, -16, 6)$, $m = 1$, and $n = 2$.

x-coordinate of A = $\frac{1(10) + 2(4)}{1 + 2} = \frac{10 + 8}{3} = \frac{18}{3} = 6$

y-coordinate of A = $\frac{1(-16) + 2(2)}{1 + 2} = \frac{-16 + 4}{3} = \frac{-12}{3} = -4$

z-coordinate of A = $\frac{1(6) + 2(-6)}{1 + 2} = \frac{6 - 12}{3} = \frac{-6}{3} = -2$

So, the coordinates of the first point A are $(6, -4, -2)$.

---

Finding the coordinates of the second point (B):

Point B divides the line segment PQ joining P$(4, 2, -6)$ and Q$(10, -16, 6)$ in the ratio $2 : 1$ internally.

Using the section formula with $m = 2$ and $n = 1$:

$B = \left(\frac{2x_2 + 1x_1}{2 + 1}, \frac{2y_2 + 1y_1}{2 + 1}, \frac{2z_2 + 1z_1}{2 + 1}\right)$

Here, $(x_1, y_1, z_1) = (4, 2, -6)$ and $(x_2, y_2, z_2) = (10, -16, 6)$.

x-coordinate of B = $\frac{2(10) + 1(4)}{2 + 1} = \frac{20 + 4}{3} = \frac{24}{3} = 8$

y-coordinate of B = $\frac{2(-16) + 1(2)}{2 + 1} = \frac{-32 + 2}{3} = \frac{-30}{3} = -10$

z-coordinate of B = $\frac{2(6) + 1(-6)}{2 + 1} = \frac{12 - 6}{3} = \frac{6}{3} = 2$

So, the coordinates of the second point B are $(8, -10, 2)$.

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Alternate method for the second point (B):

The point B is the midpoint of the line segment AQ, where A is $(6, -4, -2)$ and Q is $(10, -16, 6)$.

Using the midpoint formula for a line segment joining $(x_a, y_a, z_a)$ and $(x_b, y_b, z_b)$:

$\left(\frac{x_a + x_b}{2}, \frac{y_a + y_b}{2}, \frac{z_a + z_b}{2}\right)$

Coordinates of B = $\left(\frac{6 + 10}{2}, \frac{-4 + (-16)}{2}, \frac{-2 + 6}{2}\right) = \left(\frac{16}{2}, \frac{-20}{2}, \frac{4}{2}\right) = (8, -10, 2)$

This confirms the coordinates of the second point.

---

The coordinates of the points which trisect the line segment joining P(4, 2, -6) and Q(10, -16, 6) are (6, -4, -2) and (8, -10, 2).

Given:

The points A(1, 2, 3), B(-1, -2, -1), C(2, 3, 2), and D(4, 7, 6).

---

To Show:

1. The points A, B, C, and D are the vertices of a parallelogram ABCD.

2. The parallelogram ABCD is not a rectangle.

---

Proof:

To show that ABCD is a parallelogram, we can verify if the opposite sides are equal in length.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

---

Calculate the lengths of the sides:

Length of AB:

$AB = \sqrt{(-1 - 1)^2 + (-2 - 2)^2 + (-1 - 3)^2}$

$AB = \sqrt{(-2)^2 + (-4)^2 + (-4)^2}$

$AB = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$

Length of BC:

$BC = \sqrt{(2 - (-1))^2 + (3 - (-2))^2 + (2 - (-1))^2}$

$BC = \sqrt{(2 + 1)^2 + (3 + 2)^2 + (2 + 1)^2}$

$BC = \sqrt{(3)^2 + (5)^2 + (3)^2}$

$BC = \sqrt{9 + 25 + 9} = \sqrt{43}$

Length of CD:

$CD = \sqrt{(4 - 2)^2 + (7 - 3)^2 + (6 - 2)^2}$

$CD = \sqrt{(2)^2 + (4)^2 + (4)^2}$

$CD = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$

Length of DA:

$DA = \sqrt{(1 - 4)^2 + (2 - 7)^2 + (3 - 6)^2}$

$DA = \sqrt{(-3)^2 + (-5)^2 + (-3)^2}$

$DA = \sqrt{9 + 25 + 9} = \sqrt{43}$

---

We see that $AB = 6$ and $CD = 6$, so $AB = CD$.

We also see that $BC = \sqrt{43}$ and $DA = \sqrt{43}$, so $BC = DA$.

Since both pairs of opposite sides are equal in length, the quadrilateral ABCD is a parallelogram.

---

Check if it is a rectangle:

A parallelogram is a rectangle if and only if its diagonals are equal in length.

Calculate the lengths of the diagonals AC and BD:

Length of AC:

$AC = \sqrt{(2 - 1)^2 + (3 - 2)^2 + (2 - 3)^2}$

$AC = \sqrt{(1)^2 + (1)^2 + (-1)^2}$

$AC = \sqrt{1 + 1 + 1} = \sqrt{3}$

Length of BD:

$BD = \sqrt{(4 - (-1))^2 + (7 - (-2))^2 + (6 - (-1))^2}$

$BD = \sqrt{(4 + 1)^2 + (7 + 2)^2 + (6 + 1)^2}$

$BD = \sqrt{(5)^2 + (9)^2 + (7)^2}$

$BD = \sqrt{25 + 81 + 49} = \sqrt{155}$

---

We see that $AC = \sqrt{3}$ and $BD = \sqrt{155}$.

Since $AC \neq BD$, the lengths of the diagonals are not equal.

Therefore, the parallelogram ABCD is not a rectangle.

Hence, the statement is proved.

Let the set of points be denoted by P(x, y, z). Let the given points be A(3, 4, -5) and B(-2, 1, 4).

According to the problem, the distance from P to A is equal to the distance from P to B.

$PA = PB$

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, which is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$, we can write:

Distance PA = $\sqrt{(x - 3)^2 + (y - 4)^2 + (z - (-5))^2} \ $$ = \sqrt{(x - 3)^2 + (y - 4)^2 + (z + 5)^2}$

Distance PB = $\sqrt{(x - (-2))^2 + (y - 1)^2 + (z - 4)^2} \ $$ = \sqrt{(x + 2)^2 + (y - 1)^2 + (z - 4)^2}$

---

Since PA = PB, we have:

$\sqrt{(x - 3)^2 + (y - 4)^2 + (z + 5)^2} = \sqrt{(x + 2)^2 + (y - 1)^2 + (z - 4)^2}$

Squaring both sides to remove the square root:

$(x - 3)^2 + (y - 4)^2 + (z + 5)^2 = (x + 2)^2 + (y - 1)^2 + (z - 4)^2$

Expand the squared terms:

$(x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 + 10z + 25) = (x^2 + 4x + 4) \ $$ + (y^2 - 2y + 1) + (z^2 - 8z + 16)$

Cancel the common terms $x^2, y^2, z^2$ from both sides:

$-6x + 9 - 8y + 16 + 10z + 25 = 4x + 4 - 2y + 1 - 8z + 16$

Combine the constant terms on each side:

$9 + 16 + 25 = 50$

$4 + 1 + 16 = 21$

So, the equation becomes:

$-6x - 8y + 10z + 50 = 4x - 2y - 8z + 21$

Move all terms to one side (e.g., the left side):

$-6x - 4x - 8y + 2y + 10z + 8z + 50 - 21 = 0$

Combine like terms:

$-10x - 6y + 18z + 29 = 0$

Multiply by -1 to make the leading coefficient positive (optional but common practice):

$10x + 6y - 18z - 29 = 0$

---

The equation of the set of points P such that its distances from A(3, 4, -5) and B(-2, 1, 4) are equal is $10x + 6y - 18z - 29 = 0$. This equation represents a plane.

Given:

Coordinates of vertex A are $(x_1, y_1, z_1) = (3, -5, 7)$.

Coordinates of vertex B are $(x_2, y_2, z_2) = (-1, 7, -6)$.

Coordinates of the centroid G are $(x_G, y_G, z_G) = (1, 1, 1)$.

---

To Find:

The coordinates of the third vertex C$(x_3, y_3, z_3)$.

---

Solution:

The formula for the coordinates of the centroid of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ is:

$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)$

We are given the coordinates of A, B, and G. Let the coordinates of C be $(x_3, y_3, z_3)$.

Equating the coordinates of the centroid:

For the x-coordinate:

$x_G = \frac{x_1 + x_2 + x_3}{3}$

$1 = \frac{3 + (-1) + x_3}{3}$

$1 = \frac{2 + x_3}{3}$

$1 \times 3 = 2 + x_3$

$3 = 2 + x_3$

$x_3 = 3 - 2$

$x_3 = 1$

For the y-coordinate:

$y_G = \frac{y_1 + y_2 + y_3}{3}$

$1 = \frac{-5 + 7 + y_3}{3}$

$1 = \frac{2 + y_3}{3}$

$1 \times 3 = 2 + y_3$

$3 = 2 + y_3$

$y_3 = 3 - 2$

$y_3 = 1$

For the z-coordinate:

$z_G = \frac{z_1 + z_2 + z_3}{3}$

$1 = \frac{7 + (-6) + z_3}{3}$

$1 = \frac{1 + z_3}{3}$

$1 \times 3 = 1 + z_3$

$3 = 1 + z_3$

$z_3 = 3 - 1$

$z_3 = 2$

---

Thus, the coordinates of point C are (1, 1, 2).

Given:

Three vertices of parallelogram ABCD are A$(3, -1, 2)$, B$(1, 2, -4)$, and C$(-1, 1, 2)$.

---

To Find:

The coordinates of the fourth vertex D.

---

Solution:

Let the coordinates of the fourth vertex D be $(x, y, z)$.

In a parallelogram, the diagonals bisect each other. This means that the midpoint of the diagonal AC is the same as the midpoint of the diagonal BD.

The midpoint of a line segment joining points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula:

Midpoint $= \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$

---

Calculate the midpoint of diagonal AC:

Using A$(3, -1, 2)$ and C$(-1, 1, 2)$:

Midpoint of AC $= \left(\frac{3 + (-1)}{2}, \frac{-1 + 1}{2}, \frac{2 + 2}{2}\right)$

Midpoint of AC $= \left(\frac{2}{2}, \frac{0}{2}, \frac{4}{2}\right)$

Midpoint of AC $= (1, 0, 2)$

---

Calculate the midpoint of diagonal BD:

Using B$(1, 2, -4)$ and D$(x, y, z)$:

Midpoint of BD $= \left(\frac{1 + x}{2}, \frac{2 + y}{2}, \frac{-4 + z}{2}\right)$

---

Since the midpoints of the diagonals are the same, we equate the coordinates:

$\left(\frac{1 + x}{2}, \frac{2 + y}{2}, \frac{-4 + z}{2}\right) = (1, 0, 2)$

Equating the corresponding x-coordinates:

$\frac{1 + x}{2} = 1$

$1 + x = 2 \times 1$

$1 + x = 2$

$x = 2 - 1$

$x = 1$

Equating the corresponding y-coordinates:

$\frac{2 + y}{2} = 0$

$2 + y = 0 \times 2$

$2 + y = 0$

$y = -2$

Equating the corresponding z-coordinates:

$\frac{-4 + z}{2} = 2$

$-4 + z = 2 \times 2$

$-4 + z = 4$

$z = 4 + 4$

$z = 8$

---

Therefore, the coordinates of the fourth vertex D are (1, -2, 8).

Given:

The vertices of the triangle are A$(0, 0, 6)$, B$(0, 4, 0)$, and C$(6, 0, 0)$.

---

To Find:

The lengths of the medians of the triangle ABC.

---

Solution:

Let the midpoints of the sides BC, AC, and AB be D, E, and F respectively.

The medians of the triangle are AD, BE, and CF.

The midpoint of a line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

Midpoint $= \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$

---

1. Find the coordinates of the midpoints:

Coordinates of D (midpoint of BC):

D $= \left(\frac{0 + 6}{2}, \frac{4 + 0}{2}, \frac{0 + 0}{2}\right) = \left(\frac{6}{2}, \frac{4}{2}, \frac{0}{2}\right) = (3, 2, 0)$

Coordinates of E (midpoint of AC):

E $= \left(\frac{0 + 6}{2}, \frac{0 + 0}{2}, \frac{6 + 0}{2}\right) = \left(\frac{6}{2}, \frac{0}{2}, \frac{6}{2}\right) = (3, 0, 3)$

Coordinates of F (midpoint of AB):

F $= \left(\frac{0 + 0}{2}, \frac{0 + 4}{2}, \frac{6 + 0}{2}\right) = \left(\frac{0}{2}, \frac{4}{2}, \frac{6}{2}\right) = (0, 2, 3)$

---

2. Calculate the lengths of the medians:

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Length of median AD (joining A$(0, 0, 6)$ and D$(3, 2, 0)$):

$AD = \sqrt{(3 - 0)^2 + (2 - 0)^2 + (0 - 6)^2}$

$AD = \sqrt{(3)^2 + (2)^2 + (-6)^2}$

$AD = \sqrt{9 + 4 + 36}$

$AD = \sqrt{49} = 7$

Length of median BE (joining B$(0, 4, 0)$ and E$(3, 0, 3)$):

$BE = \sqrt{(3 - 0)^2 + (0 - 4)^2 + (3 - 0)^2}$

$BE = \sqrt{(3)^2 + (-4)^2 + (3)^2}$

$BE = \sqrt{9 + 16 + 9}$

$BE = \sqrt{34}$

Length of median CF (joining C$(6, 0, 0)$ and F$(0, 2, 3)$):

$CF = \sqrt{(0 - 6)^2 + (2 - 0)^2 + (3 - 0)^2}$

$CF = \sqrt{(-6)^2 + (2)^2 + (3)^2}$

$CF = \sqrt{36 + 4 + 9}$

$CF = \sqrt{49} = 7$

---

The lengths of the medians of the triangle are AD = 7, BE = $\sqrt{34}$, and CF = 7.

Given:

Vertices of the triangle PQR are P$(2a, 2, 6)$, Q$(-4, 3b, -10)$, and R$(8, 14, 2c)$.

The centroid of the triangle is the origin, O$(0, 0, 0)$.

---

To Find:

The values of $a$, $b$, and $c$.

---

Solution:

The formula for the coordinates of the centroid of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ is:

$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)$

In this case, $(x_1, y_1, z_1) = (2a, 2, 6)$, $(x_2, y_2, z_2) = (-4, 3b, -10)$, $(x_3, y_3, z_3) = (8, 14, 2c)$, and the centroid $G = (0, 0, 0)$.

Equating the coordinates of the centroid:

For the x-coordinate:

$0 = \frac{2a + (-4) + 8}{3}$

$0 = \frac{2a + 4}{3}$

$0 \times 3 = 2a + 4$

$0 = 2a + 4$

$2a = -4$

$a = \frac{-4}{2}$

$a = -2$

For the y-coordinate:

$0 = \frac{2 + 3b + 14}{3}$

$0 = \frac{16 + 3b}{3}$

$0 \times 3 = 16 + 3b$

$0 = 16 + 3b$

$3b = -16$

$b = -\frac{16}{3}$

For the z-coordinate:

$0 = \frac{6 + (-10) + 2c}{3}$

$0 = \frac{-4 + 2c}{3}$

$0 \times 3 = -4 + 2c$

$0 = -4 + 2c$

$2c = 4$

$c = \frac{4}{2}$

$c = 2$

---

The values are $a = -2$, $b = -\frac{16}{3}$, and $c = 2$.

Given:

The given point is P$(3, -2, 5)$.

The distance from the required point to P is $5\sqrt{2}$.

The required point lies on the y-axis.

---

To Find:

The coordinates of the point on the y-axis.

---

Solution:

Let the coordinates of the point on the y-axis be Q. Since the point lies on the y-axis, its x and z coordinates are 0.

So, let Q be $(0, y, 0)$.

The distance between Q$(0, y, 0)$ and P$(3, -2, 5)$ is given as $5\sqrt{2}$.

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

---

Substitute the coordinates of P and Q and the given distance into the formula:

$QP = \sqrt{(3 - 0)^2 + (-2 - y)^2 + (5 - 0)^2}$

$5\sqrt{2} = \sqrt{(3)^2 + (-2 - y)^2 + (5)^2}$

$5\sqrt{2} = \sqrt{9 + (-(2 + y))^2 + 25}$

$5\sqrt{2} = \sqrt{9 + (2 + y)^2 + 25}$

$5\sqrt{2} = \sqrt{34 + (2 + y)^2}$

---

Square both sides of the equation to eliminate the square root:

$(5\sqrt{2})^2 = 34 + (2 + y)^2$

$25 \times 2 = 34 + (2 + y)^2$

$50 = 34 + (2 + y)^2$

Isolate the term with y:

$(2 + y)^2 = 50 - 34$

$(2 + y)^2 = 16$

Take the square root of both sides:

$2 + y = \pm\sqrt{16}$

$2 + y = \pm 4$

---

This gives two possible cases for the value of y:

Case 1: $2 + y = 4$

$y = 4 - 2$

$y = 2$

In this case, the coordinates of the point are $(0, 2, 0)$.

Case 2: $2 + y = -4$

$y = -4 - 2$

$y = -6$

In this case, the coordinates of the point are $(0, -6, 0)$.

---

Thus, there are two points on the y-axis that are at a distance of $5\sqrt{2}$ from the point P(3, -2, 5). The coordinates of these points are (0, 2, 0) and (0, -6, 0).

Given:

Points P$(2, -3, 4)$ and Q$(8, 0, 10)$.

Point R lies on the line segment PQ and its x-coordinate is 4.

---

To Find:

The coordinates of the point R.

---

Solution:

Let the point R with coordinates $(4, y, z)$ lie on the line segment PQ.

Since R lies on the line segment PQ, it divides PQ internally in some ratio. Let this ratio be $k : 1$.

Using the section formula, the coordinates of a point that divides the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $k : 1$ are:

$\left(\frac{k x_2 + 1 x_1}{k + 1}, \frac{k y_2 + 1 y_1}{k + 1}, \frac{k z_2 + 1 z_1}{k + 1}\right)$

Here, $(x_1, y_1, z_1) = P(2, -3, 4)$, $(x_2, y_2, z_2) = Q(8, 0, 10)$, and the point of division is R$(4, y, z)$.

Equating the coordinates of R with the section formula coordinates:

For the x-coordinate:

For the y-coordinate:

$y = \frac{k(0) + 1(-3)}{k + 1} = \frac{-3}{k + 1}$

For the z-coordinate:

$z = \frac{k(10) + 1(4)}{k + 1} = \frac{10k + 4}{k + 1}$

---

Now, we solve equation (i) for $k$:

$4(k + 1) = 8k + 2$

$4k + 4 = 8k + 2$

$4 - 2 = 8k - 4k$

$2 = 4k$

$k = \frac{2}{4}$

$k = \frac{1}{2}$

Since $k$ is positive, the division is internal, which is consistent with R lying on the line segment PQ.

---

Now substitute the value of $k = \frac{1}{2}$ into the expressions for the y and z coordinates of R:

For the y-coordinate:

$y = \frac{-3}{k + 1} = \frac{-3}{\frac{1}{2} + 1} = \frac{-3}{\frac{1}{2} + \frac{2}{2}} = \frac{-3}{\frac{3}{2}} = -3 \times \frac{2}{3} = -2$

For the z-coordinate:

$z = \frac{10k + 4}{k + 1} = \frac{10(\frac{1}{2}) + 4}{\frac{1}{2} + 1} = \frac{5 + 4}{\frac{3}{2}} = \frac{9}{\frac{3}{2}} = 9 \times \frac{2}{3} = 6$

---

So, the coordinates of point R are $(4, -2, 6)$.

---

The coordinates of the point R are (4, -2, 6).

Given:

Point A is $(3, 4, 5)$.

Point B is $(-1, 3, -7)$.

The condition is $PA^2 + PB^2 = k^2$, where P is a point $(x, y, z)$ and $k$ is a constant.

---

To Find:

The equation of the set of points P.

---

Solution:

Let the coordinates of the point P be $(x, y, z)$.

The square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$

---

Calculate $PA^2$ using the points P$(x, y, z)$ and A$(3, 4, 5)$:

$PA^2 = (x - 3)^2 + (y - 4)^2 + (z - 5)^2$

$PA^2 = (x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 - 10z + 25)$

$PA^2 = x^2 - 6x + y^2 - 8y + z^2 - 10z + 9 + 16 + 25$

$PA^2 = x^2 - 6x + y^2 - 8y + z^2 - 10z + 50$

---

Calculate $PB^2$ using the points P$(x, y, z)$ and B$(-1, 3, -7)$:

$PB^2 = (x - (-1))^2 + (y - 3)^2 + (z - (-7))^2$

$PB^2 = (x + 1)^2 + (y - 3)^2 + (z + 7)^2$

$PB^2 = (x^2 + 2x + 1) + (y^2 - 6y + 9) + (z^2 + 14z + 49)$

$PB^2 = x^2 + 2x + y^2 - 6y + z^2 + 14z + 1 + 9 + 49$

$PB^2 = x^2 + 2x + y^2 - 6y + z^2 + 14z + 59$

---

Substitute these expressions for $PA^2$ and $PB^2$ into the given condition $PA^2 + PB^2 = k^2$:

$(x^2 - 6x + y^2 - 8y + z^2 - 10z + 50) + (x^2 + 2x + y^2 - 6y + z^2 \ $$ + 14z + 59) = k^2$

Group the terms with the same variables:

$(x^2 + x^2) + (-6x + 2x) + (y^2 + y^2) + (-8y - 6y) + (z^2 + z^2) + (-10z \ $$ + 14z) + (50 + 59) = k^2$

$2x^2 - 4x + 2y^2 - 14y + 2z^2 + 4z + 109 = k^2$

Rearrange the equation to put constant terms on the right side:

$2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z = k^2 - 109$

Divide the entire equation by 2:

$x^2 + y^2 + z^2 - 2x - 7y + 2z = \frac{k^2 - 109}{2}$

---

The equation of the set of points P is $x^2 + y^2 + z^2 - 2x - 7y + 2z = \frac{k^2 - 109}{2}$.

This equation represents a sphere, provided $k^2 - 109 > 0$ or $k^2 - 109 = 0$ (a single point). If $k^2 - 109 < 0$, there are no real points satisfying the condition.