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.

Given, the coordinates of point P are (a,b,c).
Which shows that, OA = a, OB = b and OC = c.
Now, point A lies on X-axis, so its coordinates are (a,0,0). Point D lies in XY-plane, so its coordinates are (a,b,0). Point B lies on Y-axis, so its coordinates are (0,b,0).
Point C lies on Z - axis, so its coordinate are (0,0,c) and point E lies in YZ-plane, so its coordinate are (0,b,c).
Hence, the coordinates of required points are
A(a,0,0), D(a,b,0), B(0,b,0), C(0,0,c) and E(0,b,c).

Coordinates of centroid of a \(\Delta ABC\) = (0,0,0)
\(\Rightarrow \left( \frac { \alpha -2+4 }{ 3 } ,\frac { 1+\beta +7 }{ 3 } ,\frac { 3-5+\gamma }{ 3 } \right) =(0,0,0)\)
\(\alpha =-2,\quad \beta =-8,\quad \gamma =2\)

\(Let\quad P(x,y,z)\quad be\quad required\quad point.\)
\(Then,\quad OP=PA=PB=PC\)
\(Now,\quad OP=PA\Rightarrow { OP }^{ 2 }={ PA }^{ 2 }\)
\(\Rightarrow { (0-x) }^{ 2 }+{ (0-y) }^{ 2 }+{ (0-z) }^{ 2 }={ x-1 }^{ 2 }+{ y-0 }^{ 2 }+{ (z-0) }^{ 2 }\)
\( \Rightarrow { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }={ x }^{ 2 }-2lx+{ l }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }\)
\(\Rightarrow 2lx={ l }^{ 2 }\Rightarrow x=\frac { 1 }{ 2 } \)
\(Similarly,\quad OP=PB\Rightarrow y=\frac { m }{ 2 } and\quad OP=PC\Rightarrow x=\frac { n }{ 2 } \)
\(Hence,\quad the\quad coordinates\quad of\quad the\quad required\quad point\quad are\quad \left( \frac { 1 }{ 2 } ,\frac { m }{ 2 } ,\frac { n }{ 2 } \right) .\)

Yes, ;et A(0, 7, 10), B(-1, 6, 6) and C(-4, 9, 6) be the vertices of a triangle. Then,
Side \(AB=\sqrt { { \left( 0+1 \right) }^{ 2 }+{ \left( 7-6 \right) }^{ 2 }+{ \left( 10-6 \right) }^{ 2 } }\)
\( [\therefore \quad distance\quad =\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-{ y }_{ 2 } \right) }^{ 2 }+{ \left( { z }_{ 1 }-{ z }_{ 2 } \right) }^{ 2 } } ]\)
\(\Rightarrow AB=\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 4 }^{ 2 } } =\sqrt { 1+1+16 } =\sqrt { 18 } =\sqrt [ 3 ]{ 2 } units\)
\(Side\quad BC=\sqrt { { \left( -1+4 \right) }^{ 2 }+{ \left( 6-9 \right) }^{ 2 }+{ \left( 6-6 \right) }^{ 2 } } =\sqrt { { 3 }^{ 2 }+{ 3 }^{ 2 }+0 } \)
\(\Rightarrow BC=\sqrt { 9+9+0 } =\sqrt { 18 } =\sqrt [ 3 ]{ 2 } units\)
\(and\quad side\quad CA=\sqrt { { \left( -4-0 \right) }^{ 2 }+{ \left( 9-7 \right) }^{ 2 }+{ \left( 6-10 \right) }^{ 2 } } \)
\(=\sqrt { { 4 }^{ 2 }+{ 2 }^{ 2 }+{ 4 }^{ 2 } } \)
\( \Rightarrow CA=\sqrt { 16+4+16 } =\sqrt { 36 } =6\quad units\)
\(Now, { AB }^{ 2 }+{ BC }^{ 2 }={ \left( \sqrt [ 3 ]{ 2 } \right) }^{ 2 }+{ \left( \sqrt [ 3 ]{ 2 } \right) }^{ 2 }{ =6 }^{ 2 }={ CA }^{ 2 }\)
Hence, \(\Delta ABC\) is right angled triangle at B.
Yes, this question gives us message to protect our environment and help us to follow these in our daily lives to make ourselvegs healthy.

Let A(x, y, z) be any point which is equidistant from points A(1, 2, 3) and B (3, 2, -1).
Then AB = \(\sqrt { { \left( x-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z-3 \right) }^{ 2 } } \)
AC = \(\sqrt { { \left( x-3 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z+1 \right) }^{ 2 } } \)
It is given that AB = AC
\(\therefore\)  \(\sqrt { { \left( x-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z-3 \right) }^{ 2 } } \)= \(\sqrt { { \left( x-3 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z+1 \right) }^{ 2 } } \)
\(\Rightarrow { \left( x-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z-3 \right) }^{ 2 }\ =\ { \left( x-3 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z+1 \right) }^{ 2 }\)
\(\Rightarrow\) x²+ 1 - 2x + z² + 9 - 6z = x² + 9 - 6x + z² + 1 + 2z
\(\Rightarrow\) - 2x - 6z + 10 = - 6x + 2z + 10
\(\Rightarrow\) -2x - 6z + 6x - 2z = 0
\(\Rightarrow\) 4x - 8z = 0
\(\Rightarrow\) x - 2z = 0.

Let the point on X-axis be P(x,0,0).
\(Then,\quad { (x-3) }^{ 2 }+{ (0-2) }^{ 2 }+{ (0-2) }^{ 2 }\)
\(={ (x-5) }^{ 2 }+{ (0-5) }^{ 2 }+{ (0-4) }^{ 2 }\)
\(Ans.\left( \frac { 49 }{ 4 } ,0,0 \right) \)

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

(1, 1, -2)

L(3,0,0), M(0,4,0) and N(0,0,5)

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

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

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

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

1:3 externally