Question

Find the equation plane passing through points $(1,2,3),(0,-1,0)$ and parallel to the line $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z}{-3}$

Answer 1

Let equation of the plane be $a(-1)+b(y-2)+c(7-3)=0$

$(0,-1,0)$

Now the putting the coordinate points we get

$-a-3b-3c=0$

$a+3b+c=0---\left(i\right)$

and

$2a+3b-3c=0$

$a+3b+3c=0------\left(ii\right)$

Now

$\frac{a}{9+9}=\frac{-b}{6+3}=\frac{c}{6-3}$

$\frac{a}{18}=\frac{b}{-9}=\frac{c}{3}$

$\frac{a}{6}=\frac{b}{-3}=\frac{c}{1}\left[ontaking3common\right]$

$6(x-1-3(y-2)+(7-3)=0$

$6x-3y+7=6-6+3$

$6x-3y+7=3$

These is the required answer.