Since,  AD is the bisector of \(\angle B A C\)
\(\Rightarrow \ \frac{B D}{D C}=\frac{A B}{A C}\) [by angle bisector theorem] ...(i)
\(\text { Now, } A B=\sqrt{(5-3)^{2}+(3-2)^{2}+(2-0)^{2}} \)
\([\because \text { distance } \left.=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\right] \)
\(=\sqrt{2^{2}+1^{2}+2^{2}}=\sqrt{4+1+4}=\sqrt{9}=3 \text { units } \)
\(\text { and } A C =\sqrt{(-9-3)^{2}+(6-2)^{2}+(-3-0)^{2}}\)
\(=\sqrt{(-12)^{2}+(4)^{2}+(-3)^{2}} \)
\(=\sqrt{144+16+9}=\sqrt{169}=13 \text { units }\)
\(Then, from Eq. (i), \frac{B D}{D C}=\frac{3}{13}\)
Hence, D divides BC in the ratio 3 : 13.Then, coordinates of D are
\(\left[\frac{3(-9)+13(5)}{3+13}, \frac{3(6)+13(3)}{3+13}, \frac{3(-3)+13(2)}{3+13}\right]\) [using internal ratio formula]
\(=\left(\frac{-27+65}{16}, \frac{18+39}{16}, \frac{-9+26}{16}\right)=\left(\frac{38}{16}, \frac{57}{16}, \frac{17}{16}\right)=\left(\frac{19}{8}, \frac{57}{16}, \frac{17}{16}\right)\)

Let C (0, \(\frac{1}{3}\), 2) divides the joint of A (2, -3) and B(-1, 2,1) in the ratio k:1
Then coordinates of C are 
\(\left( \frac { -k+2 }{ k+1 } ,\frac { 2k-3 }{ k+1 } ,\frac { k+4 }{ k+1 } \right) \) [using internal ratio formula]
But coordinates of C are  (0, \(\frac{1}{3}\), 2)
On comparing Eqs.(i) and (ii) we get
\(\frac { -k+2 }{ k+1 } =0\Rightarrow -k+2=0\Rightarrow 2\)
\( \frac { 2k-3 }{ k+1 } =\frac { 1 }{ 3 } \Rightarrow 6k-9=k+1\Rightarrow 5k=10\Rightarrow k=2\)
\(and \quad \frac { k+4 }{ k+1 } =2\Rightarrow k+4=2k+2\Rightarrow k=2\)
From each of these equations, we get k=2 Since, from each equation, we get the same value of k. Therefore, the given points are collinear nad C divides AB internally in the ratio 2:1.

CGLet ABCD be a tetrahedron, such that the coordinates of its vertices are A(x₁,y₁,z₁),B(x₂,y₂,z₂),C(x₃,y_3,z₃) and D(x₄,y₄,z₄), respectively.

Let the centroid of the faces BCD,ACD,ABD,ABC be G₁,G₂,G₃ and G₄ respectively.Then centroid  G₁ of face BCD
\(=\left( \frac { { x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 } }{ 3 } ,\frac { { y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 } }{ 3 } ,\frac { z_{ 2 }+{ z }_{ 3 }+{ z }_{ 4 } }{ 3 } \right) \)
Now, coordinates of point G dividing AG₁ in the ratio 3:1 are
\(\left[ \frac { 1.{ x }_{ 1 }+3\left( { \frac { { x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 } }{ 3 } } \right) }{ 1+3 } \frac { 1.{ y }_{ 1 }+3\left( \frac { { y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 } }{ 3 } \right) }{ 1+3 } \frac { 1.{ z }_{ 1 }+3\left( \frac { { z }_{ 2 }+{ z }_{ 3 }+{ z }_{ 4 } }{ 3 } \right) }{ 1+3 } \right] \)
\( =\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 } }{ 4 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 } }{ 4 } ,\frac { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 }+{ z }_{ 4 } }{ 4 } \right) \)
Similarly, the point dividing BG₂ ,CG₃ and DG₄ in the ration 3:1 has same coordinates.Thus, the point
\(\\ G\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 } }{ 4 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 } }{ 4 } ,\frac { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 }+{ z }_{ 4 } }{ 4 } \right) \)
is common AG₁,BG_2, CG_3,DG₄

\({ CA }^{ 2 }+{ AB }^{ 2 }\neq { BC }^{ 2 }\)
Ans.No.

The vertices of the triangle are A(1, 2, 3), B(3, 4, 5) and C(-1, 6, 7). Also, centroid of the triangles is G(1, 4, 1/3).

Let Q(O,y, 0) be any point on y-axis. Then
\(PQ=\sqrt { (0-3)2+(y+2)2+(0-5)2 } \)
\(=\sqrt { 9+y2+4+4y+25 } \)
\(=\sqrt { y2+4y+38 } \)
But \(\sqrt { y2+4y+38 } =5\sqrt { 2 } \)
y² + 4y + 38 = 50 \(\Rightarrow \\ \) y² + 4y - 12 = 0 \(\Rightarrow \\ \) (y - 2) (y + 6) = 0
\(\Rightarrow \\ \) y = 2, -6
Thus coordinates of point Q are (0, 2, 0) and (0, -6,0).

Here A (x₁, y₁, z₁) B (x_2, y_2, z₂₎and C (x_3, y₃, z₃ )be three vertices of \(\triangle \)ABC, then coordinates of point D are
\(\left[ \frac { x1+x2+x3 }{ 2 } ,\frac { y1+y2+y3 }{ 2 } ,\frac { z1+z2+z3 }{ 2 } \right] \)

Let Gbe the centroid of ABC. Then Gdivides AD in the ratio 2 : 1. So the coordinates of G are
\(\left[ \frac { x1+2\left( \frac { x2+x3 }{ 2 } \right) }{ 1+2 } ,\frac { y1+2\left( \frac { y2+y3 }{ 2 } \right) }{ 1+2 } ,\frac { z1+2\left( \frac { z2+z3 }{ 2 } \right) }{ 1+2 } \right] \)
\(\Rightarrow \left( \frac { x1+x2+x3 }{ 3 } ,\frac { y1+y2+y3 }{ 3 } ,\frac { z1+z2+z3 }{ 3 } \right) \)

Here A (3a, 4, -5), B (-2, 4b, 6), and C (6, 10, c) be three vertices of \(\triangle \)ABC, then coordinates of centroid are 
\(\left[ \frac { 3a-2+6 }{ 3 } ,\frac { 4+4b+10 }{ 3 } ,\frac { -5+6+c }{ 3 } \right] \)
But it is given that co-ordinates of centroid are (0, 0, 0).
\(\therefore \frac { 3a-2+6 }{ 3 } =0\Rightarrow 3a=-4.\Rightarrow a=-\frac { 4 }{ 3 } \\ \frac { 4+4b+10 }{ 3 } =0\Rightarrow 4b=-14\Rightarrow -\frac { 7 }{ 2 } \\ \frac { -5+6+c }{ 3 } =0\Rightarrow c=-1\)

Let Q divides PR in the ratio k : l. Thus co-ordinates of Q are
\(\left[ \frac { 11k+5 }{ k+1 } ,\frac { 10k+4 }{ k+1 } ,\frac { -8k-2 }{ k+1 } \right] \)
It is given that coordinates of Q are (7, 6, - 4).
\(\therefore \frac { 11k+5 }{ k+1 } =7,\frac { 10k+4 }{ k+1 } =6,\frac { -8k-2 }{ k+1 } =-4\)
Now solving these, we get k=\(\frac { 1 }{ 2 } \) .
Thus Q divides PR in the ratio \(\frac { 1 }{ 2 } \) : 1 or 1 : 2.

Let P(x, y, z) be any point which is equidistant from A(3, 4, 0) and B(5, 2, -3).
Now PA = PB => PA² = PB²
\(\therefore \)(x - 3)² + (y - 4)² + (z - 0)²
= (x - 5)² + (y - 2)² + (z + 3)²
=> x²+ 9 - 6x + y² + 16 - 8y + Z²
= x² + 25 -10x + y² + 4 - 4y + Z²+ 9 + 6z
=> 4x - 4y - 6z - 13 = O.

Let A(2, 0, 3), B(O, 3, 2) and C(O, 0,1) be given points.
Let P(x, y, 0) be any point in XY plane such that PA = PB = PC.
Now PA = PB => PA2 = PB2
\(\therefore \) (x - 2)² + (y - 0)² + (- 3)²
= (x - 0)² + (y - 3)² + (- 2)²
=> x²+4-4x+y²+9
=x²+y²+9-6y+4
=> 4x - 6y = 0 => 2x - 3y = 0 ....(i)
Also PB = PC => PB² = PC²
(x - 0)² + (y - 3)² + (0 - 2)² = (x - 0)² + (y - 3)² + (0 - 2)²
\(\therefore \)(x - 0)² + (y - 3)² + (0 - 2)² = (x - 0)² + (y - 3)² + (0 - 1)²
=> x²+ y² + 9 - 6y + 4
=x²+y²+1
=> 6y = 12 => y = 2
Putting value of y in (i), we have
2x-3x2=0 => x=3
Thus co-ordinates ofrequired point are (3, 2, 0).