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

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

The correct answer is option (4) → $7x-8y+3z+25=0$

$\vec{n_1}⊥x+2y+3z=5⇒\vec{n_1}=\hat i+2\hat j+3\hat k$

$\vec{n_2}⊥3x+3y+z=0⇒\vec{n_2}=3\hat i+3\hat j+\hat k$

so $\vec n$ ⊥ given plane passing through (-1, 3, 2) = $\vec{n_1}×\vec{n_2}$

$\vec n=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&3\\3&3&1\end{vmatrix}=-7\hat i+8\hat j-3\hat k$

so $(\vec r-\vec a).\vec n=0$

$\vec a = -\hat i+3\hat j+2\hat k$

$\vec r.\vec n=\vec a.\vec n$

$-7x+8y-3z=+7+24-6$

so $7x-8y+3z+25=0$