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
 A: Hint: Any plane passing through intersection of $4x - 2y + z - 3 = 0$ and $2x - y + 3z + 1 = 0$ is given by $$(4x - 2y + z - 3) + k(2x - y + 3z + 1) = 0$$ or what is the same as $$(2k + 4)x - (k + 2)y + (3k + 1)z + k - 3 = 0$$
This is perpendicular to $3x + y - z + 7 = 0$. Using dot product of normal vectors you can now find $k$.
EDIT: If you do calculations you will find $k = -9/2$ and final answer would be same as that provided in another answer namely $-2x + y - 5z = 3$ or $2x - y + 5z + 3 = 0$
A: 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 $$-2x+y-5z = 3.$$
A: The answer $-2x + y - 5z = 3$  is the right answer. The simple way to go about it is to first find a vector which lies on the plane. This vector is parallel to the line of intersection of the two planes.  We therefore find the cross product of the two normals to the intersecting planes. We get the vector $-5i - 10j$  this is then crossed with the normal to the third plane to which  the required plane is perpendicular. This gives us the normal to the required plane which is $2i - j + 5k$.                                                       Then a point on the plane is obtained from the equations of the intersecting planes. Set say $z=0$. Solve simultaneously for $x$ and $y$.    Thank you.
A: Here is a different approach which gives us the same answer in the end. Let:
Let Plane P1 be: 4x-2y+z=3 with normal vector $\vec{n_1}$=[4 -2 1] 
$$ $$
Plane P2: 2x-y+3z=-1 has $\vec{n_2}$=[2 -1 3] as its normal vector.
$$ $$
Plane P3: 3x+y-z=-7 has $\vec{n_3}$=[3 1 -1] as its normal vector.
$$ $$
Plane P4: is the plane that we are seeking.
$$ $$
To find the line of intersection of P1 with P2, we can perform gaussian on those two planes:  \begin{bmatrix}2&-1&3&-1\\4&-2&1&3\end{bmatrix} 
which will give us a solution of the form: $$\begin{bmatrix}x&\\y&\\z\end{bmatrix} =\begin{bmatrix}1&\\0&\\-1\end{bmatrix}+
t\begin{bmatrix}1&\\2&\\0\end{bmatrix} \qquad[1]$$ where t $\epsilon$ R. 
Equation [1] is the line of intersection of P1 with P2 and has a direction vector equal $$\vec{v}= \begin{bmatrix}1&\\2&\\0\end{bmatrix}$$. To get the equation of the plane P4, we first find its normal vector by taking the following cross product:
$$ \vec{n_4}=\vec{n_3}\times \vec{v}=\begin{bmatrix}\vec{i}&\vec{j}&\vec{k}\\1&2&0\\3&1&-1\end{bmatrix} =-2i+j-5k$$
So $\vec{n_4}$ is the normal vector of P4. 
$$ $$
Next, we have to find a point that lies on P4. Since P4 passes through the line of intersection of P1 and P2, P4 contains the line of intersection give by equation [1] above. So any points on the line of intersection will do. Indeed, pt (3,4,-1) as you found lies on the line of intersection (this can also be achieved by letting t=2 in equation [1] above)  and so does (1,0,-1). Suppose we choose (1,0,-1), then the equation of the plane P4 is:
$$-2x+y-5z=3$$
which is the same answer as given by others above as expected.
