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 = \(\sqrt { { \left( 7+2 \right) }^{ 2 }+{ \left( 0-3 \right) }^{ 2 }+{ \left( -1-5 \right) }^{ 2 } } \)
\(\sqrt { 81+9+36 } \)
\(=\sqrt { 126 } \)
\(=3\sqrt { 14 } \)
\(\text { Here, } \mathrm{PQ}+\mathrm{QR}=\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}=\mathrm{PR}\)
\(\text { Hence, points } \mathrm{P}(-2,3,5), \mathrm{Q}(1,2,3), \text { and } \mathrm{R}(7,0,-1) \text { are collinear. }\)

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.

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 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
\(P Q =\sqrt{(-4-1)^{2}+(1+3)^{2}+(2-4)^{2}} \)
\(=\sqrt{25+16+4} \)
\(=\sqrt{45}=3 \sqrt{5} \text { units }\)