Question

Find the equation of the plane passes through $(-1,3,2)$ and perpendicular to each of the two planes. $x+2y+2z=5,3x+3y+2z=8$

Answer 1

Required plane is perpendicular to both planes $x+2y+2z=5$(having dirction ration $n_1$) and $3x+3y+2z=8$ (having direction ration $n_2$) so perpendicular plane will have direction rations $n_1\times n_2=-2i+4j-3k$

equation of plane will be $-2x+4y-3z=c$

This plane contain $(-1,3,2)$ so $-2(-1)+4\left(3\right)-3\left(2\right)=c$

$c=8$

Equation of plane $-2x+4y-3z=8$