Let Rand S be two point which trisect the join of PQ.
\(\therefore \) PR=RS= SQ

\(\therefore \) Point R divides the join of PQ in the ratio 1:2
\(\therefore \) Coordinates of R is
\(\left( \frac { 1\times 10+2\times 4 }{ 1+2 } ,\frac { 2\times -16+1\times 2 }{ 1+2 } ,\frac { 2\times 6+1\times -6 }{ 1+2 } \right) \)
= (6, -4, -2).
Also point S divides the join of PQ in the ratio 2:1
\(\therefore \) Coordinates of S is
\(\left( \frac { 2\times 10+1\times 4 }{ 1+2 } ,\frac { 2\times -16+1\times 2 }{ 1+2 } ,\frac { 2\times 6+1\times -6 }{ 1+2 } \right) \)
= (8, - 10, 2).
\(\text { Thus, }(6,-4,-2) \text { and }(8,-10,2) \text { are the points that trisect the line segment joining }\)\(\text { points } P(4,2,-6) \text { and } Q(10,-16,6)\)

Let Q(5, 4, - 6) divides the line segment joining points P(3, 2, -4) and R(9, 8, -10) in the ratio k : 1 internally.
\(\therefore\) Then coordinates of Q are
\(\left( \frac { 9k+3 }{ k+1 } ,\frac { 8k+2 }{ k+1 } ,\frac { -10k-4 }{ k+1 } \right) \)
But it is given that coordinates of Q is (5, 4, - 6)
\(\therefore\) \(\frac { 9k+3 }{ k+1 } =5\) \(\Rightarrow\) 9k + 3 = 5k + 5
\(\Rightarrow\) 4k = 2 \(\Rightarrow\) k = \(\frac { 1 }{ 2 } \)
Thus Q divides the line segment joining points P and R in the ratio \(\frac { 1 }{ 2 } \) : 1 i.e. 1 : 2 internally.

Let P(x, y, z) be any point which divides the line segment joining points A (-2, 3, 5) and B (1, -4, 6) in the ratio 2 : 3 externally.
Then
\(x=\frac { 2\times 1+\left( -3 \right) \times -2 }{ 2+\left( -3 \right) } =\frac { 2+6 }{ -1 } =-8\)
\(y=\frac { 2\times -4+\left( -3 \right) \times 3 }{ 2+\left( -3 \right) } =\frac { -8-9 }{ -1 } =17\)
\(z=\frac { 2\times 6+\left( -3 \right) \times 5 }{ 2+\left( -3 \right) } =\frac { 12-15 }{ -1 } =3\)
\(\therefore\) Coordinates of P are (-8, 17,3).

Let the line joining the points A(1, 2, 3) and B(-3, 4, - 5) is divided by the xy-plane at a point C in the ratio k : 1.
Then coordinates of C are
\(x=\frac { -3k+1 }{ k+1 } ,\ y=\frac { 4k+2 }{ k+1 } ,\ z=\frac { -5k+3 }{ k+1 } \)
Since the point, C lies on XY-plane. So z = 0.
\(\therefore\)  \(\frac { -5k+3 }{ k+1 } =0\Rightarrow \) -5k + 3 = 0 \(\Rightarrow\)k=\(\frac { 3 }{ 5 } \)
Thus the required ratio is \(\frac { 3 }{ 5 } \):1 i.e. 3: 5.

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)

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

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

Let A (-1, 2, 1), B (1, -2, 5) and C(4, -7, 8) and D(2, -3, 4) be four vertices of a quadrilateral ABCD. Then
AB = \(\sqrt { { \left( 1+1 \right) }^{ 2 }+{ \left( -2-2 \right) }^{ 2 }+{ \left( 5-1 \right) }^{ 2 } } \)
= \(\sqrt { 4+16+16 } =\sqrt { 36 } =6\)
BC = \(\sqrt { { \left( 4-1 \right) }^{ 2 }+{ \left( -7+2 \right) }^{ 2 }+{ \left( 8-5 \right) }^{ 2 } } \)
\(\sqrt { 9+25+9 } =\sqrt { 43 } \)
CD = \(\sqrt { { \left( 2-4 \right) }^{ 2 }+{ \left( -3+7 \right) }^{ 2 }+{ \left( 4-8 \right) }^{ 2 } } \)
\(\sqrt { 4+16+16 } =\sqrt { 36 } =6\)
AD = \(\sqrt { { \left( 2+1 \right) }^{ 2 }+{ \left( -3-2 \right) }^{ 2 }+{ \left( 4-1 \right) }^{ 2 } } \)
= \(\sqrt { 9+25+9 } =\sqrt { 43 } \)
AC = \(\sqrt { { \left( 4+1 \right) }^{ 2 }+{ \left( -7-2 \right) }^{ 2 }+{ \left( 8-1 \right) }^{ 2 } } \)
\(\sqrt { 25+81+49 } =\sqrt { 155 } \)
BD = \(\sqrt { { \left( 2-1 \right) }^{ 2 }+{ \left( -3+2 \right) }^{ 2 }+{ \left( 4-5 \right) }^{ 2 } } \)
\(\sqrt { 1+1+1 } =\sqrt { 3 } \)
Now AB = CD, BC = AD and AC\(\neq\)BD
Thus A, B, C, and D are vertices of a parallelogram ABCD.

Let A(0, 7, 10), B(-I, 6, 6) and C(- 4,9,6) be three vertices of triangle ABC. Then
AB = \(\sqrt { { \left( -1-0 \right) }^{ 2 }+{ \left( 6-7 \right) }^{ 2 }+{ \left( 6-10 \right) }^{ 2 } } \)
\(\sqrt { 1+1+16 } =\sqrt { 18 } =3\sqrt { 2 } \)
BC = \(\sqrt { { \left( -4+1 \right) }^{ 2 }+{ \left( 9-6 \right) }^{ 2 }+{ \left( 6-6 \right) }^{ 2 } } \)
\(\sqrt { 9+9+0 } =\sqrt { 18 } =3\sqrt { 2 } \)
AC = \(\sqrt { { \left( -4-0 \right) }^{ 2 }+{ \left( 9-7 \right) }^{ 2 }+{ \left( 6-10 \right) }^{ 2 } } \)
\(\sqrt { 16+4+16 } =\sqrt { 36 } =6\)
Now AC² = AB² + BC²
Thus, ABC is a right angled triangle.

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

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

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

Given points are (x,-8,4) and (3,-5,4) and distance between these points = 5
\(\therefore \sqrt { { (x-3) }^{ 2 }+{ { (-8+5) }^{ 2 } }+{ (4-4) }^{ 2 } } =5\)
\(\Rightarrow \sqrt { { (x-3) }^{ 2 }+{ { (-3) }^{ 2 } }+0 } =5\)
\(\Rightarrow\)(x - 3)² + 9 + 0 = 25 [ on squaring both sides]
\(\Rightarrow\)(x - 3)² = 16 \(\Rightarrow\) (x - 3)² = (4)²
\(\Rightarrow\) x - 3 = \(\pm \)4 [ taking square root on both sides]
\(\therefore \)  x = 7 or  -1

Since, in XY-plane, z-coordinate will be zero.Hence, the coordinates of the foot of the perpendicular L(5,4,0).

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