Let P(x, y, z) be any point which divides the line segment joining points A(3, -2, 5) and B(3, 4, 2) in the ratio 2 : 1 externally.
Then,
\(x=\frac { 2\times 3+\left( -1 \right) \times 3 }{ 2+\left( -1 \right) } =\frac { 6-3 }{ 1 } =3\)
\(y=\frac { 2\times 4+\left( -1 \right) \times -2 }{ 2+\left( -1 \right) } =\frac { 8+2 }{ 1 } =10\)
\(z=\frac { 2\times 2+\left( -1 \right) \times 5 }{ 2+\left( -1 \right) } =\frac { 4-5 }{ 1 } =-1\)
\(\therefore\)   Coordinates of P are (3, 10, - 1)

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

3 units

Let A (- 1, 3, - 4) and B(1, - 3, 4) be two points. Then
AB = \(\sqrt { { \left( 1-\left( -1 \right) \right) }^{ 2 }+{ \left( -3-3 \right) }^{ 2 }+{ \left( 4-\left( -4 \right) \right) }^{ 2 } } \)
\(=\sqrt { 4+36+64 } \)
\(=\sqrt { 104 } \)
\(=2\sqrt { 26 } \)

Here AB = \(\sqrt { { \left( -1-1 \right) }^{ 2 }+{ \left( -1-2 \right) }^{ 2 }+{ \left( -1-3 \right) }^{ 2 } } =\sqrt { 4+9+16 } =\sqrt { 29 } \)
BC = \(\sqrt { { \left( 3-\left( -1 \right) \right) }^{ 2 }+{ \left( 5-\left( -1 \right) \right) }^{ 2 }+{ \left( 7-\left( -1 \right) \right) }^{ 2 } } \)
= \(\sqrt { { \left( 3+1 \right) }^{ 2 }+{ \left( 5+1 \right) }^{ 2 }+{ \left( 7+1 \right) }^{ 2 } } =\sqrt { 16+36+64 } =\sqrt { 116 } =2\sqrt { 29 } \)
AC = \(\sqrt { { \left( 3-1 \right) }^{ 2 }+{ \left( 5-2 \right) }^{ 2 }+{ \left( 7-3 \right) }^{ 2 } } =\sqrt { 4+9+16 } =\sqrt { 29 } \)
Now BC = AB + AC
Thus, A, B, and C are collinear points.

Given point is (2,3,4). Here, x-coordinate i.e. 2 is positive. So, we take a point A in positive direction of X -axis at a distance 2 from O. Thus, point A be (2,0,0). From point A, we drw a line perpendicular to X-axis which will be parallel to Y-axis and take point B at a distance 3 from A in positive direction.Thus, point B be (2,3,0).Draw a line prallel to Z-axis from point B, which is perpendicular to XY-plane and take point C at this line in positive direction at a distance 4 units from B. Thus, point C(2,3,4) is the required location of given point in space, which is shown in the figure.

L is the foot of perpendicular drwn from the point P(3,4,5) to the XY-plane.
Therefore, the coordinate of the point L are (3,4,0). The distance between the points (3,4,5) and (3,4,0) is 5 units. Similarly, the lengths of the foot of perpendiculars on YZ and ZX-planes are 3 and 4 units, respectively.

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

\(Given\quad points\quad are\quad A(3,4,5)\quad and\quad B(-1,3,-7).\)
\(Let\quad the\quad coordinates\quad of\quad point\quad P\quad be\quad (x,y,z).\)
\(Then, { PA }^{ 2 }={ (x-3) }^{ 2 }+{ (y-4) }^{ 2 }+{ (z-5) }^{ 2 }\)
\( \left[ \because distance\quad =\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 }+{ ({ z }_{ 2 }-{ z }_{ 1 }) }^{ 2 } } \right] \)
\(and \quad { PB }^{ 2 }={ (x+1) }^{ 2 }+{ (y-3) }^{ 2 }+{ (z+7) }^{ 2 }\)
\(By\quad the\quad given\quad condition\quad { PA }^{ 2 }+{ PB }^{ 2 }={ 2k }^{ 2 },\)
\( { (x-3) }^{ 2 }+{ (y-4) }^{ 2 }+{ (z-5) }^{ 2 }+{ (x+1) }^{ 2 }+{ (y-3) }^{ 2 }+{ (z+7) }^{ 2 }={ 2k }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+9-6x+{ y }^{ 2 }+16-8y+{ z }^{ 2 }+25-10z\)
\(+{ x }^{ 2 }+2x+1+{ y }^{ 2 }+9-6y+{ z }^{ 2 }+49+14z={ 2k }^{ 2 }\)
\(\Rightarrow { 2x }^{ 2 }+{ 2y }^{ 2 }+{ 2z }^{ 2 }-4x-14y+4z={ 2k }^{ 2 }-109\)
\(which\quad is\quad the\quad required\quad equation.\)

The centroid of a triangle is equal to the centroid of the triangle formed by mid-points of its sides.

(1, 1, -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).

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
\(\left[ \frac { x1+2\left( \frac { x2+x3 }{ 2 } \right) }{ 1+2 } ,\frac { y1+2\left( \frac { y2+y3 }{ 2 } \right) }{ 1+2 } ,\frac { z1+2\left( \frac { z2+z3 }{ 2 } \right) }{ 1+2 } \right] \)
\(\Rightarrow \left( \frac { x1+x2+x3 }{ 3 } ,\frac { y1+y2+y3 }{ 3 } ,\frac { z1+z2+z3 }{ 3 } \right) \)

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.