Question

Prove that the points $(1,2,3),(3,0,3),(-2,-3,-3)$ and $(3,4,6)$ are coplanar

Answer 1

Equation to plane passing through the point $(1,2,3)$
$a(x-1)+b(y-2)+z(z-3)=\mathrm{0.....}\left(1\right)$
$\because$ plane (1) passes through point $(3,0,3)$ and $(-2,-3,-3)$
$\therefore$ $a(3-1)+b(0-2)+c(3-3)=0$
$2a-2b+0c=\mathrm{2.....}\left(2\right)$
and
$a(-2-1)+b(-3-2)+c(-3-3)=0$
$3a+5b+6c=\mathrm{0.....}\left(3\right)$
from equaion (2) and (3)
$\frac{a}{-12+0}=\frac{b}{0-12}=\frac{c}{10+6}$
$\frac{a}{-12}=\frac{b}{-12}=\frac{c}{16}$
$\frac{a}{3}=\frac{b}{3}=\frac{c}{-4}$
Let $\frac{a}{3}=\frac{b}{3}=\frac{c}{-4}=K$
$a=3K,b=3K,c=-4K$
putting the value of $a,b,c$ in equation (1)
$3K(x-1)+3K(y-2)-4K(z-3)=0$
$K\left[3\right(x-1)+3(y-2)-4(z-3\left)\right]=0$
$3x+3y-4z+3=0$
LHS$=3x+3y-4z+3$
$=3\left(3\right)+3\left(4\right)-4\left(6\right)+3$
$24-24$
$=0$
Hence the given points are coplanar