Find the equation of the plane that passes through the line of intersection of the planes... Find the equation of the plane that passes through the line of intersection of the planes $4x - 2y + z - 3 = 0$ and $2x - y + 3z + 1 = 0$, and that is perpendicular to the plane $3x + y - z + 7 = 0$.
This is what I got: $3x + 4y - z + 15 = 0$.
Can you please tell me if this is right?
Thanks in advance!
Here is my work:

EDIT:
Changed my answer to: 2x + 3y + 9z - 9 = 0

Hint: The line of intersection of the 2 planes is parallel to
$$ \begin{pmatrix} 4 \\ -2 \\ 1 \\ \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ 3 \\ \end{pmatrix} = \begin{pmatrix} -5 \\ -10  \\ 0 \end{pmatrix}.$$
Hint: The plane that you are interested in is parallel to $\begin{pmatrix} 1 \\ 2  \\ 0 \end{pmatrix}$ and parallel to $\begin{pmatrix}3 \\ 1 \\ -1 \end{pmatrix}$. Hence, it also is perpendicular to 
$$ \begin{pmatrix} 1 \\ 2 \\ 0 \\ \end{pmatrix} \times \begin{pmatrix} 3 \\ 1 \\ -1 \\ \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\-5 \end{pmatrix}.$$
Hint: You found that the point $(3,4,-1)$ lies on both planes, hence lies on the line of intersection, hence lies on the plane that you are interested in.
Thus, the equation of the plane is $$-2x+y-5z = 3.$$
