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 $x+2y–3z=8$.

Solution The equation of the plane containing the given point is
$A(x–1)+B(y+1)+C(z–2)=0...\left(1\right)$
Applying the condition of perpendicularity to the plane given in (1) with the planes
$2x+3y–2z=5$ and $x+2y–3z=8$, we have
$2A+3B–2C=0$ and $A+2B–3C=0$
Solving these equations, we find $A=–5C$ and $B=4C$. Hence, the required equation is
$–5C(x–1)+4C(y+1)+C(z–2)=0$
$i.e5x–4y–z=7$