Let A(-2,6,-2), B(0,4,-1), C(-2,3,1) and D (-4,5,0) be the given points.


\(\therefore B=\sqrt { { (0+2) }^{ 2 }+{ (4-6 })^{ 2 }+{ (1+2) }^{ 2 } } \)   [Using the distance formula]
\(=\sqrt { 4+4+1 } =\sqrt { 9 } =3\quad units\)
\(BC=\sqrt { { (-2-0) }^{ 2 }+{ (3-4) }^{ 2 }+({ 1+1) }^{ 2 } } \)
\(\sqrt { 4+1+4 } =\sqrt { 9 } =3\quad units\)
\( CD=\sqrt { { (-4+2) }^{ 2 }+{ (5-3) }^{ 2 }+{ (0-1) }^{ 2 } } \)
\(=\sqrt { 4+4+1 } =\sqrt { 9 } =3\quad units\)
\(AD=\sqrt { (-4+{ 2) }^{ 2 }+{ { (5-6) } }^{ 2 }+({ 0+2) }^{ 2 } } \)
\(=\sqrt { 4+1+4 } =\sqrt { 9 } =3\quad units\)
Here, AB=BC=CD=DA
Now, \(AC=\sqrt { { (-2+2) }^{ 2 }+{ (3-6) }^{ 2 }+({ 1+2) }^{ 2 } } \)
\([\because \quad distance=\sqrt { ({ x }_{ 2 }-{ x }_{ 1 }{ ) }^{ 2 }+{ (y }_{ 2 }-{ y }_{ 1 }{ ) }^{ 2 }+{ (z }_{ 2 }-{ z }_{ 1 }{ ) }^{ 2 } } \)
\(=\sqrt { 0+9+9=\sqrt { 18\quad units } } \)
\(and\quad BD=\sqrt { ({ -4 }-0{ ) }^{ 2 }+5-4{ ) }^{ 2 }+(0-1{ ) }^{ 2 } } \)
\(=\sqrt { 16+1+1 } =\sqrt { 18 } units\)
Sine,diagonal AC=diagonal BD
Hence, ABCD is a square.

Let A(0,7,10), B(-1,6,6) and C(-4,9,6) be the given points.
Then AB= \(=\sqrt { { (-1- }0)^{ 2 }+({ 6-7) }^{ 2 }+(6-{ 10 })^{ 2 } } \)  [using the distance formula]
\(=\sqrt { 1+1+16 } =\sqrt { 18 } =3\sqrt { 2 } \)units

\(BC=\sqrt { { (-4+1 })^{ 2 }+{ (9- }6)^{ 2 }+{ (6-6 })^{ 2 } } \)
\(=\sqrt { 9+9+0 } =\sqrt { 18 } =3\sqrt { 2 } units\)
\(and\quad AC=\sqrt { { (-4-0) }^{ 2 }+{ (9-7) }^{ 2 }+{ (6-10) }^{ 2 } } =\sqrt { 16+4+16 }\)
\( \Rightarrow AC=\sqrt { 36 } =6\quad units\)
\(Now,\quad { AB }^{ 2 }+{ BC }^{ 2 }=(3\sqrt { 2 } { ) }^{ 2 }+(3\sqrt { 2 } { ) }^{ 2 }=18+18=36\quad units\)
\(\therefore { AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }\)
\( Also,\quad AB={ BC }^{ 2 }={ AC }^{ 2 }\)
Hence, ABC is a right angled isosceles triangle.

Let P(x,y,z) be any point on the given curve and let
A(3,4,−5) and B(−2,1,4) be the given points.
Then,PA=PB⇒PA²=PB²
⇒(x−3)²+(y−4)²+(z+5)²=(x+2)²+(y−1)²+(z−4)²
⇒x²+9−6x+y²+16−8y+z²+25+10z
=x²+4+4x+y²+1−2y+z²+16−8z
⇒10x+6y−18z−29=0
Hence,the required curve is
10x+6y−18z−29=0

Show that AD=BD=CD

Show that AB = BC = CA

When we draw a perpendicular line from the point P(3,6,9) on the  YZ-plane, the x-coordinate of foot of perpendicular will be zero and the other coordinates ( y and z) will be 6 and 9, i.e coordinates of a point on YZ-plane (which is the foot of perpendicular drawn from P to YZ plane) be Q(0,6,9).
\(\therefore \) Distance between P and Q,
\(QP=\sqrt { { (3-0) }^{ 2 }+{ (6-6) }^{ 2 }+{ (9-9) }^{ 2 } } \)
\([\because distance=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 }+{ { (z }_{ 2 }-{ z }_{ 1 }) }^{ 2 } } \)
\(=\sqrt { { 3 }^{ 2 }+{ 0 }^{ 2 }+{ 0 }^{ 2 } } \)
= 3 Units

2:1 internally

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

If a point is on the x-axis, then its y-coordinates and z-coordinates are zero.

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

Let third vertex of \(\Delta ABC\) be C(x,y,z).
We know that intersection of medians of triangle is known as centroid of a triangle.
\(\therefore \left( \frac { 3+1+x }{ 3 } ,\frac { 4+3+y }{ 3 } \frac { 2+2+z }{ 3 } \right) =(2,4,3)\)
C(2,5,5)

\(\left( \frac { -3 }{ 10 } ,\frac { 5 }{ 2 } ,\frac { -1 }{ 5 } \right) \)

Let P(x, y, z) be any point Then
\(PA=\sqrt { (x-3)^{ 2 }+(y-4)^{ 2 }+(z-5)^{ 2 } } \)
\(=\sqrt { x^{ 2 }+9-6x+y^{ 2 }+16-8y+z^{ 2 }+25-10z } \)
\(PA=\sqrt { (x+1)^{ 2 }+(y-3)^{ 2 }+(z+7)^{ 2 } } \)
\(\sqrt { x^{ 2 }+1+2x+y^{ 2 }+9-6y+z^{ 2 }+49-14z }\)
Now PA² + PB² = K²

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).