Find the equations of the line passing through the point $(3,1,-2)$ and parallel to each of the planes $x-y+z=4$ and $3x+y-z=5$.
I started off by finding the normal vector from both of the planes $ \begin{vmatrix} i & j & k \\ 1 & -1 & 1 \\ 3 & 1 & -1 \end{vmatrix} $ which worked out to $<0,4,4>$, but at this point I'm not sure where to continue.
On
Hint:
A directing vector of the sought line has to be orthogonal to the normal vectors of each plane, hence has to be collinear to the cross-product $\;\vec n_1\times \vec n_2$.
Thus a parametric equation of the line parallel to both planes, passing through point $A\,(3,1,-2)$ will be $$ M=A+t\,(\vec n_1\times \vec n_2). $$
You have found the vector $\vec n=(0,4,4)^T$ that is orthogonal to the orthogonal of the first plane and orthogonal to the orthogonal of the second plane, so it is parallel to the two planes, and the line is $$ \vec x=P+t\cdot\vec n $$
where $P=(3,1,-2)$ is the given point.
So we have: $$ \begin{pmatrix} x\\y\\z \end{pmatrix} = \begin{pmatrix} 3\\1+4t\\-2+4t \end{pmatrix} $$ that is equivalent to: $$ \begin{cases} x=3\\ y=1+4t\\ z=-2+4t \end{cases} $$ and, subtracting the last two equation we find: $$ \begin{cases} x=3\\ y-z=1+2\\ \end{cases} $$ and that is another form of the equation of the line.