Question

Find the equation of the plane that contains the point $(1,-1,2)$ and is perpendicular to each of the planes $2x+3y-2z=5$ and $x+2y-3z=8$.

Answer 1

The equation of a plane passing through $(1,-1,2)$ is $a(x-1)+b(y+1)+c(z-2)=0$
It is perpendicular to the planes $2x+3y-2z=5$ and $x+2y-3z=8$.
Therefore, $2a+3b-2c=0$ and, $a+2b-3c=0$
Solving these two equation by cross-multiplication, we get $\frac{a}{-5}=\frac{b}{4}=\frac{c}{1}$
Substituting $a=-5c,b=4c$ and $c=1$ in (i),
we get $-5x+4y+z=-7$ as the equation of the required plane.