From the Table , the point (–3,1, 2) lies in second octant and the point (–3, 1, – 2) lies in octant VI.

For the point F, the distance measured along OY is zero. Therefore, the coordinates of F are (2,0,5).

Let A(- 2, 3, 5), B(1, 2, 3) and C(7, 0, -1) be three given points.
Then PQ = \(\sqrt { { \left( 1+2 \right) }^{ 2 }+{ \left( 2-3 \right) }^{ 2 }+{ \left( 3-5 \right) }^{ 2 } } \)
= \(\sqrt { 9+1+4 }\)
\( =\sqrt { 14 } \)
QR = \(\sqrt { { \left( 7-1 \right) }^{ 2 }+{ \left( 0-2 \right) }^{ 2 }+{ \left( -1-3 \right) }^{ 2 } } \)
\(\sqrt { 36+4+16 } \)
\(=\sqrt { 56 }\)
\( =2\sqrt { 14 } \)
PR = \(\sqrt { { \left( 7+2 \right) }^{ 2 }+{ \left( 0-3 \right) }^{ 2 }+{ \left( -1-5 \right) }^{ 2 } } \)
\(\sqrt { 81+9+36 } \)
\(=\sqrt { 126 } \)
\(=3\sqrt { 14 } \)
\(\text { Here, } \mathrm{PQ}+\mathrm{QR}=\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}=\mathrm{PR}\)
\(\text { Hence, points } \mathrm{P}(-2,3,5), \mathrm{Q}(1,2,3), \text { and } \mathrm{R}(7,0,-1) \text { are collinear. }\)

Let A(x, y, z) be any point which is equidistant from points A(1, 2, 3) and B (3, 2, -1).
Then AB = \(\sqrt { { \left( x-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z-3 \right) }^{ 2 } } \)
AC = \(\sqrt { { \left( x-3 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z+1 \right) }^{ 2 } } \)
It is given that AB = AC
\(\therefore\)  \(\sqrt { { \left( x-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z-3 \right) }^{ 2 } } \)= \(\sqrt { { \left( x-3 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z+1 \right) }^{ 2 } } \)
\(\Rightarrow { \left( x-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z-3 \right) }^{ 2 }\ =\ { \left( x-3 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z+1 \right) }^{ 2 }\)
\(\Rightarrow\) x²+ 1 - 2x + z² + 9 - 6z = x² + 9 - 6x + z² + 1 + 2z
\(\Rightarrow\) - 2x - 6z + 10 = - 6x + 2z + 10
\(\Rightarrow\) -2x - 6z + 6x - 2z = 0
\(\Rightarrow\) 4x - 8z = 0
\(\Rightarrow\) x - 2z = 0.

Let P(x, y, z) be any point.
Then PA = \(\sqrt { { \left( x-4 \right) }^{ 2 }+{ \left( y-0 \right) }^{ 2 }+{ \left( z-0 \right) }^{ 2 } } =\sqrt { { x }^{ 2 }+16-8x+{ y }^{ 2 }+{ z }^{ 2 } } \)
PB = \(\sqrt { { \left( x+4 \right) }^{ 2 }+{ \left( y-0 \right) }^{ 2 }+{ \left( z-0 \right) }^{ 2 } } =\sqrt { { x }^{ 2 }+16+8x+{ y }^{ 2 }+{ z }^{ 2 } } \)
It is given that PA + PB = 10
\(\therefore\) \(\sqrt { { x }^{ 2 }+16-8x+{ y }^{ 2 }+{ z }^{ 2 } } +\sqrt { { x }^{ 2 }+16+8x+{ y }^{ 2 }+{ z }^{ 2 } } =10\)
\(\Rightarrow \sqrt { { x }^{ 2 }+16-8x+{ y }^{ 2 }+{ z }^{ 2 } } =10-\sqrt { { x }^{ 2 }+16+8x+{ y }^{ 2 }+{ z }^{ 2 } } \)
Squaring both sides, we have
x² + 16 - 8x + y² + Z² = 100 + x² + 16 + 8x + y² + Z² - 20\(\sqrt { { x }^{ 2 }+16+8x+{ y }^{ 2 }+{ z }^{ 2 } } \)
\(\Rightarrow 20\sqrt { { x }^{ 2 }+16+8x+{ y }^{ 2 }+{ z }^{ 2 } } \) = 16x + 100
\(\Rightarrow 5\sqrt { { x }^{ 2 }+16+8x+{ y }^{ 2 }+{ z }^{ 2 } } \) = 4x + 25
Squaring both sides again, we have
25(x² + 16 + 8x + y² + z²) = 16x² + 625 + 200x
\(\Rightarrow\) 25x² + 400 + 200x + 25y² + 25z² -16x² - 625 - 200x = 0
\(\Rightarrow\) 9x² + 25y² + 25z² - 225 = 0
Thus the required equation is 9x² + 25y² + 25z² - 225 = 0

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

If a point is on the x-axis, then its y-coordinates and z-coordinates are zero.

The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I.
The x-coordinate, y-coordinate, and z-coordinate of point (4, –2, 3) are positive, negative, and positive respectively. Therefore, this point lies in octant IV.
The x-coordinate, y-coordinate, and z-coordinate of point (4, –2, –5) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.
The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, –5) are positive, positive, and negative respectively. Therefore, this point lies in octant V.
The x-coordinate, y-coordinate, and z-coordinate of point (–4, 2, –5) are negative, positive, and negative respectively. Therefore, this point lies in octant VI.
The x-coordinate, y-coordinate, and z-coordinate of point (–4, 2, 5) are negative, positive, and positive respectively. Therefore, this point lies in octant II.
The x-coordinate, y-coordinate, and z-coordinate of point (–3, –1, 6) are negative, negative, and positive respectively. Therefore, this point lies in octant III.
The x-coordinate, y-coordinate, and z-coordinate of point (2, –4, –7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

Let P(x, y, z) be any point Then
\(PA=\sqrt { (x-3)^{ 2 }+(y-4)^{ 2 }+(z-5)^{ 2 } } \)
\(=\sqrt { x^{ 2 }+9-6x+y^{ 2 }+16-8y+z^{ 2 }+25-10z } \)
\(PA=\sqrt { (x+1)^{ 2 }+(y-3)^{ 2 }+(z+7)^{ 2 } } \)
\(\sqrt { x^{ 2 }+1+2x+y^{ 2 }+9-6y+z^{ 2 }+49-14z }\)
Now PA² + PB² = K²

Here P(2a, 2, 6), Q(-4, 3b, -10) and R(8, 14, 2c) are vertices of triangle PQR.
\(\therefore \)Coordinates of centroid of PQR is
\(\left( \frac { 2a-4+8 }{ 3 } ,\frac { 2+3b+14 }{ 3 } ,\frac { 6-10+2c }{ 3 } \right) \)
\(=\left( \frac { 2a+4 }{ 3 } ,\frac { 6-10+2c }{ 3 } ,\frac { 2c-4 }{ 3 } \right) \)
But is it given that coordinates of centroid is (0,0,0)
\(\therefore \frac { 2a+4 }{ 3 } =0\Rightarrow 2a+4=0\Rightarrow a+-2\)
\( \frac { 3b+16 }{ 3 } =0\Rightarrow 3b+16=0\)
\(\Rightarrow b=\frac { -16 }{ 3 } \)
\(\frac { 2c-4 }{ 3 } =0\Rightarrow 2c-4=0\Rightarrow c=2.\)

Here A(0,0, 6), B(0,4, 0) and C(6, 0, 0) are vertices of \(\triangle \)ABC

Now D is mid point of BC 
\(\therefore \) Coordinates of D is \(\left( \frac { 0+6 }{ 2 } ,\frac { 4+0 }{ 2 } ,\frac { 0+0 }{ 2 } \right) \)= (3,2,0)
\(\therefore \quad AD=\sqrt { (0-3)^{ 2 }+(0-2)^{ 2 }+(6-0)^{ 2 } } \)
=\(\sqrt { 9+4+36 } =\sqrt { 7 } \) units.
Also E is mid point of AC
\(\therefore \)  Coordinates of E is \(\left( \frac { 0+6 }{ 2 } ,\frac { 0+0 }{ 2 } ,\frac { 6+0 }{ 2 } \right) \)= (3,0,3)
\(\therefore \quad BE=\sqrt { (0-3)^{ 2 }+(4-0)^{ 2 }+(0-3)^{ 3 } } \)
\(=\sqrt { 9+16+9 } =\sqrt { 34 } \)units.
Also F is mid point of AB
\(\therefore \) Coordiates of F is \(\left( \frac { 0+0 }{ 2 } ,\frac { 0+4 }{ 2 } ,\frac { 6+0 }{ 2 } \right) \)=(0,2,3)
\(\therefore \) CF=\(\sqrt { (6-0)^{ 2 }+(0-2)^{ 2 }+(0-3)^{ 2 } } \)
\(=\sqrt { 36+4+9 } =7\)units.

Let D(x, y, z) be the fourth vertex of parallelogram ABCD.
We know that diagonals of a parallelogram bisect each other. So the mid points ofAC and BD coincide.

\(\therefore \) Coordinates of mid point of AC \(\left( \frac { 3-1 }{ 2 } ,\frac { -1+1 }{ 2 } ,\frac { 2+2 }{ 2 } \right) \)=(1,0,2)
Also coordiantes of mid point of BD \(\left( \frac { x+1 }{ 2 } ,\frac { y+2 }{ 2 } ,\frac { z-4 }{ 2 } \right) \)
\(\therefore \quad \frac { x+1 }{ 2 } =1\Rightarrow x+1=2\Rightarrow x=1\)
\(\frac { y+2 }{ 2 } =0\Rightarrow y+2=0\Rightarrow y=-2\)
\(\frac { z-4 }{ 2 } =2\Rightarrow z-4=4\Rightarrow z=8\)
Thus the coordinates of point Dare (1, -2,8).

(i) Let A(0, 7, -10), B(1, 6, -6) and C(4, 9, -6) be three vertices of triangle ABC.Then
AB = \(\sqrt { { \left( 1-0 \right) }^{ 2 }+{ \left( 6-7 \right) }^{ 2 }+{ \left( -6+10 \right) }^{ 2 } } \)
\(\sqrt { 1+1+16 } =\sqrt { 18 } =3\sqrt { 2 } \)
BC = \(\sqrt { { \left( 4-1 \right) }^{ 2 }+{ \left( 9-6 \right) }^{ 2 }+{ \left( -6+6 \right) }^{ 2 } } \)
\(\sqrt { 9+9+0 } =\sqrt { 18 } =3\sqrt { 2 } \)
AC = \(\sqrt { { \left( 4-0 \right) }^{ 2 }+{ \left( 9-7 \right) }^{ 2 }+{ \left( -6+10 \right) }^{ 2 } } \)
\(\sqrt { 16+4+16 } =\sqrt { 36 } =6\)
Now AB = BC
Thus, ABC is an isosceles triangle.
(ii) Let A(0, 7, 10), B(-I, 6, 6) and C(- 4,9,6) be three vertices of triangle ABC. Then
AB = \(\sqrt { { \left( -1-0 \right) }^{ 2 }+{ \left( 6-7 \right) }^{ 2 }+{ \left( 6-10 \right) }^{ 2 } } \)
\(\sqrt { 1+1+16 } =\sqrt { 18 } =3\sqrt { 2 } \)
BC = \(\sqrt { { \left( -4+1 \right) }^{ 2 }+{ \left( 9-6 \right) }^{ 2 }+{ \left( 6-6 \right) }^{ 2 } } \)
\(\sqrt { 9+9+0 } =\sqrt { 18 } =3\sqrt { 2 } \)
AC = \(\sqrt { { \left( -4-0 \right) }^{ 2 }+{ \left( 9-7 \right) }^{ 2 }+{ \left( 6-10 \right) }^{ 2 } } \)
\(\sqrt { 16+4+16 } =\sqrt { 36 } =6\)
Now AC² = AB² + BC²
Thus, ABC is a right angled triangle.
(iii) Let A (-1, 2, 1), B (1, -2, 5) and C(4, -7, 8) and D(2, -3, 4) be four vertices of a quadrilateral ABCD. Then
AB = \(\sqrt { { \left( 1+1 \right) }^{ 2 }+{ \left( -2-2 \right) }^{ 2 }+{ \left( 5-1 \right) }^{ 2 } } \)
= \(\sqrt { 4+16+16 } =\sqrt { 36 } =6\)
BC = \(\sqrt { { \left( 4-1 \right) }^{ 2 }+{ \left( -7+2 \right) }^{ 2 }+{ \left( 8-5 \right) }^{ 2 } } \)
\(\sqrt { 9+25+9 } =\sqrt { 43 } \)
CD = \(\sqrt { { \left( 2-4 \right) }^{ 2 }+{ \left( -3+7 \right) }^{ 2 }+{ \left( 4-8 \right) }^{ 2 } } \)
\(\sqrt { 4+16+16 } =\sqrt { 36 } =6\)
AD = \(\sqrt { { \left( 2+1 \right) }^{ 2 }+{ \left( -3-2 \right) }^{ 2 }+{ \left( 4-1 \right) }^{ 2 } } \)
= \(\sqrt { 9+25+9 } =\sqrt { 43 } \)
AC = \(\sqrt { { \left( 4+1 \right) }^{ 2 }+{ \left( -7-2 \right) }^{ 2 }+{ \left( 8-1 \right) }^{ 2 } } \)
\(\sqrt { 25+81+49 } =\sqrt { 155 } \)
BD = \(\sqrt { { \left( 2-1 \right) }^{ 2 }+{ \left( -3+2 \right) }^{ 2 }+{ \left( 4-5 \right) }^{ 2 } } \)
\(\sqrt { 1+1+1 } =\sqrt { 3 } \)
Now AB = CD, BC = AD and AC\(\neq\)BD
Thus A, B, C, and D are vertices of a parallelogram ABCD.

(i) Let A (2, 3, 5) and B(4, 3, 1) be two points.
\(\mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\)
Then
AB = \(\sqrt { { \left( 4-2 \right) }^{ 2 }+{ \left( 3-3 \right) }^{ 2 }+{ \left( 1-5 \right) }^{ 2 } } \)
= \(\sqrt { 4+0+16 } \)
= \(\sqrt { 20 } \)
= \(2\sqrt { 5 } \) units
(ii) Let A (- 3, 7, 2) and B(2, 4, - 1) be two points. Then
AB = \(\sqrt { { \left( 2-\left( -3 \right) \right) }^{ 2 }+{ \left( 4-7 \right) }^{ 2 }+{ \left( -1-2 \right) }^{ 2 } } \)
= \(\sqrt { { \left( 2+3 \right) }^{ 2 }+{ \left( 4-7 \right) }^{ 2 }+{ \left( -1-2 \right) }^{ 2 } } \)
= \(\sqrt { 25+9+9 } \)
= \(\sqrt { 43 } \) units
(iii) Let A (- 1, 3, - 4) and B(1, - 3, 4) be two points. Then
AB = \(\sqrt { { \left( 1-\left( -1 \right) \right) }^{ 2 }+{ \left( -3-3 \right) }^{ 2 }+{ \left( 4-\left( -4 \right) \right) }^{ 2 } } \)
\(=\sqrt { 4+36+64 } \)
\(=\sqrt { 104 } \)
\(=2\sqrt { 26 } \) units
(iv) Let A (2, - 1, 3) and B(- 2, 1, 3) be two points. Then
AB = \(\sqrt { { \left( -2-2 \right) }^{ 2 }+{ \left( 1-\left( -1 \right) \right) }^{ 2 }+{ \left( 3-3 \right) }^{ 2 } } \)
= \(\sqrt { { \left( -2-2 \right) }^{ 2 }+{ \left( 1+1 \right) }^{ 2 }+{ \left( 3-3 \right) }^{ 2 } } \)
\(\sqrt { 16+4+0 }\) 
\( =\sqrt { 20 } =2\sqrt { 5 } \)units.

Let the coordinates of C be (x, y, z) and the coordinates of the centroid G be (1, 1, 1). Then
\(\frac{x+3-1}{3}=1 \text { , i.e., } x=1 ; \frac{y-5+7}{3}=1 \text { , i.e., } y=1 ; \frac{z+7-6}{3}=1 \text { , i.e., } z=2 \text { . }\)
Hence, coordinates of C are (1, 1, 2).

If P (x, y, z) be any point such that PA = PB.
\(\text { Now } \sqrt{(x-3)^{2}+(y-4)^{2}+(z+5)^{2}}=\sqrt{(x+2)^{2}+(y-1)^{2}+(z-4)^{2}}\)
\(\text { or }(x-3)^{2}+(y-4)^{2}+(z+5)^{2}=(x+2)^{2}+(y-1)^{2}+(z-4)^{2}\)
or 10 x + 6y – 18z – 29 = 0.

To show ABCD is a parallelogram we need to show opposite side are equal
Note that 
\(\mathrm{AB}=\sqrt{(-1-1)^{2}+(-2-2)^{2}+(-1-3)^{2}}=\sqrt{4+16+16}=6 \)
\(\mathrm{BC}=\sqrt{(2+1)^{2}+(3+2)^{2}+(2+1)^{2}}=\sqrt{9+25+9}=\sqrt{43} \)
\(\mathrm{CD}=\sqrt{(4-2)^{2}+(7-3)^{2}+(6-2)^{2}}=\sqrt{4+16+16}=6 \)
\(\mathrm{DA}=\sqrt{(1-4)^{2}+(2-7)^{2}+(3-6)^{2}}=\sqrt{9+25+9}=\sqrt{43}\)
Since AB = CD and BC = AD, ABCD is a parallelogram.
Now, it is required to prove that ABCD is not a rectangle. For this, we show that diagonals AC and BD are unequal. We have
\(\mathrm{AC}=\sqrt{(2-1)^{2}+(3-2)^{2}+(2-3)^{2}}=\sqrt{1+1+1}=\sqrt{3} \)
\(\mathrm{BD}=\sqrt{(4+1)^{2}+(7+2)^{2}+(6+1)^{2}}=\sqrt{25+81+49}=\sqrt{155} .\)
Since AC \(\ne\) BD, ABCD is not a rectangle

Let the coordinates of point P be (x, y, z).
Here PA² = (x – 3)² + (y – 4)² + ( z – 5)²
PB² = (x + 1)² + (y – 3)² + (z + 7)²
By the given condition PA² + PB² = 2k², we have
(x – 3)² + (y – 4)² + (z – 5)² + (x + 1)² + (y – 3)² + (z + 7)² = 2k²
i.e., 2x² + 2y² + 2z² – 4x – 14y + 4z = 2k² – 109.

By the distance formula, we have
AB² = (10 – 3)² + (20 – 6)² + (30 – 9)²
= 49 + 196 + 441 = 686
BC² = (25 – 10)² + (– 41 – 20)² + (5 – 30)²
= 225 + 3721 + 625 = 4571
CA² = (3 – 25)² + (6 + 41)² + (9 – 5)²
= 484 + 2209 + 16 = 2709
We find that CA² + AB² \(\ne\) BC².
Hence, the triangle ABC is not a right angled triangle.

We know that points are said to be collinear if they lie on a line.
\(\text { Now, }\mathrm{PQ}=\sqrt{(1+2)^{2}+(2-3)^{2}+(3-5)^{2}}=\sqrt{9+1+4}=\sqrt{14}\)
\(\mathrm{QR}=\sqrt{(7-1)^{2}+(0-2)^{2}+(-1-3)^{2}}=\sqrt{36+4+16}=\sqrt{56}=2 \sqrt{14} \)
\(\text { and } \quad P R=\sqrt{(7+2)^{2}+(0-3)^{2}+(-1-5)^{2}}=\sqrt{81+9+36}=\sqrt{126}=3 \sqrt{14}\)
Thus, PQ + QR = PR. Hence, P, Q and R are collinear

The distance PQ between the points P (1,–3, 4) and Q (– 4, 1, 2) is
\(P Q =\sqrt{(-4-1)^{2}+(1+3)^{2}+(2-4)^{2}} \)
\(=\sqrt{25+16+4} \)
\(=\sqrt{45}=3 \sqrt{5} \text { units }\)