# Find the equation of a plane that is perpendicular to another plane, parallel to a line and goes through a point

Find the equation of a plane which is perpendicular to the plane $$\pi_1\equiv x-3y-z+1=0,$$ parallel to a line $$l\equiv\frac{x - 2}{2} = \frac{y -3}{-3} = \frac{z}{1}$$ and goes through point $P = (-1, 1, 2)$.

All I know it that the normal vector of the given plane is $\overrightarrow{n_1}=(1,-3,-1)$ and the direction vector of the given line is $\overrightarrow{d_l}=(2,-3,1)$.

The plane you are looking for contains the normal vector to $\pi_1$, and also contains the direction vector of $l$. As you point out, those are $\langle 1, -3, -1\rangle$ and $\langle 2, -3, 1\rangle$. Can you use that information to figure out a normal vector to the plane you are looking for, then put that together with the given information about $P$ to find your answer?