\(\left(\frac{7}{2}, \frac{13}{2}, 9\right)\)

Let vertices of the \(\Delta\)ABC be A (x₁, y₁, z₁) B (x₂, y₂, z₂) and C (x₃, y₃, z₃) and mid-points of the sides BC, CA and AB are D(5, 7, 11),E(O, 8, 5)and F(2, 3, -1), respectively.
Now, D is the mid-point of BC.
\(\therefore \quad 5=\frac{x_{2}+x_{3}}{2} \Rightarrow x_{2}+x_{3}=10\)
\(7=\frac{y_{2}+y_{3}}{2} \Rightarrow y_{2}+y_{3}=14\)
\(\text { and } \quad 11=\frac{z_{2}+z_{3}}{2} \Rightarrow z_{2}+z_{3}=22\)
Similarly, for the sides AB and AC,
\(2 =\frac{x_{1}+x_{2}}{2} \Rightarrow x_{1}+x_{2}=4 \)
\(3 =\frac{y_{1}+y_{2}}{2} \Rightarrow y_{1}+y_{2}=6 \)
\(-1 =\frac{z_{1}+z_{2}}{2} \Rightarrow z_{1}+z_{2}=-2 \)
\(0 =\frac{x_{1}+x_{3}}{2} \Rightarrow x_{1}+x_{3}=0 \)
\(8 =\frac{y_{1}+y_{3}}{2} \Rightarrow y_{1}+y_{3}=16 \)
\(5 =\frac{z_{1}+z_{3}}{2} \Rightarrow z_{1}+z_{3}=10\)
On adding Eqs. (i) and (iv),we get
\(x_{1}+2 x_{2}+x_{3} =14 \)
\(\Rightarrow 2 x_{2}+0 =14 [\text { from Eq]. }\)
\(\Rightarrow x_{2}=7\)
On putting x₂ = 7 in Eqs. (i) and Eqs. (iv),we get
\(x_{3}=3 \text { and } x_{1}=-3\)
\(\text { Thus, } x_{1}=-3, x_{2}=7 \text { and } x_{3}=3\)
On adding Eqs. (ii) and (v), we get
\(y_{1}+2 y_{2}+y_{3}=20 \)
\(\Rightarrow 2 y_{2}+16 =20 \quad[\text { from Eq. (viii)] }\)
\(\Rightarrow 2 y_{2}=4 \Rightarrow y_{2}=2 \)
On putting y₂= 2in Eqs. (ii) and Eqs. (v),we get
\(y_{3}= 12 \text { and } y_{1}=4 \)
\(\text { Thus, } y_{1}=4, y_{2} =2 \text { and } y_{3}=12\)
On adding Eqs. (iii) and (vi), we get
\( z_{1}+2 z_{2}+z_{3} =20 \)
\(\Rightarrow 2 z_{2}+10 =20 \text { [from Eq. (ix)] }\)
\(\Rightarrow \lambda z_{2}=10 \Rightarrow z_{2}=5\)
On putting Z₂ = 5 in Eqs. (iii) and (vi), we get
\(z_{3}=17 \text { and } z_{1}=-7 \)
\(\text { Thus, } z_{1}=-7, z_{2}=5 \text { and } z_{3}=17\)
Hence, the vertices are A (- 3,4, - 7), B( 7, 2, 5) and C(3, 12,17).

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.

Let YZ-plane divides the line segment joining A (4, 8, 10) and B (6, 10, – 8) at P (x, y, z) in the ratio k : 1. Then the coordinates of P are
\(\left(\frac{4+6 k}{k+1}, \frac{8+10 k}{k+1}, \frac{10-8 k}{k+1}\right)\)
Since P lies on the YZ-plane, its x-coordinate is zero, i.e., \(\frac{4+6 k}{k+1}=0\)
\(\begin{array}{l} \text { Or } k=-\frac{2}{3} \end{array}\)
Therefore, YZ-plane divides AB externally in the ratio 2 : 3.

Let A (– 4, 6, 10), B (2, 4, 6) and C(14, 0, – 2) be the given points. Let the point P divides AB in the ratio k : 1. Then coordinates of the point P are
\(\left(\frac{2 k-4}{k+1}, \frac{4 k+6}{k+1}, \frac{6 k+10}{k+1}\right)\)
Let us examine whether for some value of k, the point P coincides with point C.
\(\text { On putting } \frac{2 k-4}{k+1}=14, \text { we get } k=-\frac{3}{2}\)
\(\text { When } k=-\frac{3}{2}, \text { then } \frac{4 k+6}{k+1}=\frac{4\left(-\frac{3}{2}\right)+6}{-\frac{3}{2}+1}=0\)
\(\text { and }\frac{6 k+10}{k+1}=\frac{6\left(-\frac{3}{2}\right)+10}{-\frac{3}{2}+1}=-2\)
Therefore, C (14, 0, –2) is a point which divides AB externally in the ratio 3 : 2 and is same as P.Hence A, B, C are collinear.

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.

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

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  A(-1, 2, 1), B(1, -2, 5), C(4, -7, 8) and D(2, -3, 4) be the vertices of a quadrilateral ABCD.

\(Then\quad the\quad mid\quad point\quad AC\)
\(=\left( \frac { -1+4 }{ 2 } ,\frac { 2-7 }{ 2 } ,\frac { 1+8 }{ 2 } \right) =\left( \frac { -3 }{ 2 } ,\frac { -5 }{ 2 } ,\frac { 9 }{ 2 } \right) \)
\(\left[ \because coordinates\quad of\quad mid-point=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { { y }_{ 1 }+ }{ y }_{ 2 } }{ 2 } ,\frac { { z }_{ 1 }+{ z }_{ 2 } }{ 2 } \right) \right] \)
\(and\quad mid\quad point\quad BD=\left( \frac { 1+2 }{ 2 } ,\frac { -2-3 }{ 2 } ,\frac { 5+4 }{ 2 } \right)\)
\(=\left( \frac { 3 }{ 2 } ,\frac { -5 }{ 2 } ,\frac { 9 }{ 2 } \right) \)
Here, mid-point of both the diagonals are same i.e they bisect each other. Hence ABCD is a parallelogram

(1, 2, 3), (3, 4, 5) and (-1, 6, -7)

Show that AB=BC=CD=DA AND AC \(\neq \) BD

\(Show\quad that\quad { AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }\)

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).

\(\left( \frac { 23 }{ 8 } ,\frac { 3 }{ 2 } ,\frac { 19 }{ 8 } \right) , \frac { \sqrt { 1530 } }{ 8 } \)

Let A (3, - 2, 4), B (1,0, - 2) and C (-1,2, -8) be the given points.
\(\text { Then, } A B=\sqrt{(1-3)^{2}+(0+2)^{2}+(-2-4)^{2}}\)  [using distance formula]
\(=\sqrt{4+4+36}=\sqrt{44}=2 \sqrt{11} \text { units }\)
\(B C=\sqrt{(-1-1)^{2}+(2-0)^{2}+(-8+2)^{2}}\) [using distance formula]
\(=\sqrt{4+4+36} \)
\(=\sqrt{44}=2 \sqrt{11} \text { units }\)
\(\text { and } C A=\sqrt{(3+1)^{2}+(-2-2)^{2}+(4+8)^{2}}\) [using distance formula]
\(=\sqrt{16+16+144} \)
\(=\sqrt{176}=4 \sqrt{11} \text { units }\)
We observe that,
\(A B+B C =2 \sqrt{11}+2 \sqrt{11} \)
\(=4 \sqrt{11}=C A\)
Hence, A, Band C are collinear.

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.

(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.