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
On
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 $2x - y + 5z + 3 = 0$
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 $$-2x+y-5z = 3.$$