\(\left( \frac { 7 }{ 3 } ,2,1 \right) \) and \(\left( \frac { 5 }{ 3 } ,2,3 \right) \)

(6,3,16)

10x + 4y - 10z - 14 = 0

6x - 8y + 16z - 55 = 0

 \(3\sqrt { 3 } \) units

6 units

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

Here AB = \(\sqrt { { \left( 2-0 \right) }^{ 2 }+{ \left( -1-1 \right) }^{ 2 }+{ \left( 3-2 \right) }^{ 2 } } =\sqrt { 4+4+1 } =3\)
BC = \(\sqrt { { \left( 1-2 \right) }^{ 2 }+{ \left( -3+1 \right) }^{ 2 }+{ \left( 1-3 \right) }^{ 2 } } =\sqrt { 1+4+4 } =3\)
AC = \(\sqrt { { \left( 1-0 \right) }^{ 2 }+{ \left( -3-1 \right) }^{ 2 }+{ \left( 1-2 \right) }^{ 2 } } =\sqrt { 1+16+1 } =\sqrt { 18 } =3\sqrt { 2 } \)
Now AB = BC and AC²= AB²+ BC²
Thus, ABC is an isosceles right-angled triangle.

Show that AB=BC=CA

Show that AB = BC = CD = DA and AC = BD

Let P(x,y,z) be the required point. According to question, AP+BP=K, where k be any arbitrary constant
\(\Rightarrow \sqrt { { \left( x-0 \right) }^{ 2 }+{ \left( y-0 \right) }^{ 2 }+{ \left( z+\propto \right) }^{ 2 } } +\sqrt { { \left( x-0 \right) }^{ 2 }+{ \left( y-0 \right) }^{ 2 }+{ \left( z+\propto \right) }^{ 2 } } =K\)
\({ 4k }^{ 2 }{ x }^{ 2 }+{ 4k }^{ 2 }{ y }^{ 2 }+{ 4 }{ z }^{ 2 }({ K }^{ 2 }-{ 4\propto }^{ 2 })\quad +\quad K^{ 2 }({ 4\propto }^{ 2 }-{ K }^{ 2 })=0\)

2:1 internally

Show that mid-point of AC is equal to the mid-point of BD.

Let C divides the join of A and B in the ratio k:1.
\(\therefore\) Coordinates of \(C=\left( \frac { -k+2 }{ k+1 } ,\frac { 2k+3 }{ k+1 } ,\frac { -3k+4 }{ k+1 } \right) \)
Equate\(\left( \frac { -k+2 }{ k+1 } ,\frac { 2k+3 }{ k+1 } ,\frac { -3k+4 }{ k+1 } \right) =(-4,\quad 1,\quad -10)\) to find ratio.
2:1 externally

(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.