Find the equation of the plane passing through the point $( - 1,3,2)$ and perpendicular to each of the planes $x + 2y + 3z = 5$ and $3x + 3y + z = 0$.

.: Any plane passing through $( - 1,3,2)$is
$a(x + 1) + b(y - 3) + c(z - 2) = 0$ ...(1)

Since plane (1) is perpendicular to the planes :
$x + 2y + 3z = 5$

and $3x + 3y + z = 0$

$\therefore$ $(a)(1) + b(2) + c(3) = 0$

and $a(3) + b(3) + c(1) = 0$

$\Rightarrow$ $a + 2b + 3c = 0$ ... (2)

and $3a + 3b + c = 0$ ...(3)

Solving (2) and (3),

we get
$\cfrac{a}{{2 - 9}} = \cfrac{b}{{9 - 1}} = \cfrac{c}{{3 - 6}} \Rightarrow \cfrac{a}{{ - 7}} = \cfrac{b}{8} = \cfrac{c}{{ - 3}} = k$(say)

So, $a = - 7k,b = 8k,c = - 3k$
Putting in (1),

we get
$- 7k(x + 1) + 8k(y - 3) - 3k(z - 2) = 0$
$\Rightarrow$ $- 7x - 7 + 8y - 24 - 3z + 6 = 0$

$\Rightarrow$ $7x - 8y + 3z + 25 = 0$,

which is the required equation.