Question

Find the equation of the plane through the points $(2,2,1)$ and $(9,3,6)$ and perpendicular to the plane $2x+6y+6z=9$.

Answer 1

Plane passing through $(2,2,1)$ and $(9,3,6)$

$a(x-2)+b(y-2)+c(z-1)=0$ - plane through $(2,2,1)$

Though it also passes through $(9,3,6)$

$7a+b+5c=0$ ……..$\left(1\right)$

It is perpendicular to $2x+6y+6z=9$

$2a+6b+6c=0$ ……$\left(2\right)$

$\frac{a}{-26}=\frac{b}{-18}=\frac{c}{40}=t$

$a=-26t$

$b=-18t$

$c=40t$

$t(-26(x-2)+(-18\left)\right(y-2)+40(2-1\left)\right)=0$

$13x+9y-20z-24=0$

$\downarrow$

Equation of plane.