Equation of the line passing through the origin and parallel to the planes $x+y+z=-1$ and $x-y+z=1$ 
Find a vector equation of the line that passes through the origin and is parallel to the planes $x+y+z=-1, x-y+z=1$

Is the answer $2x-2z=0$? I took the normals of the two planes which are $(1, 1, 1)$ and $(1, -1, 1)$ and used the cross-product to get the normal of the new plane, which is $(2, 0, -2)$. Since the line passes the origin, I would get $2x-2z=0$. Is this the right approach?
 A: Your analysis is correct.
Here is a different way to describe the line parallel to the intersection of the two planes.  Just row reduce the corresponding homogeneous system of equations.  (These planes are parallel to your given ones but passing through the origin.)
$$
\left\{
\begin{align}
x + y + z &= 0 \\
x - y + z &= 0
\end{align}
\right.
$$
As matrices,
$$
\begin{bmatrix}
1 & 1 & 1 \\
1 & -1 & 1
\end{bmatrix}
\leadsto
\begin{bmatrix}
1 & 1 & 1 \\
0 & -2 & 0
\end{bmatrix}
\leadsto
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 0
\end{bmatrix}.
$$
We have
$$
\left\{
\begin{align}
x + z &= 0 \\
y &= 0
\end{align}
\right.,
$$
which is (by renaming the free variable $z = t$) equivalent to
$$
\vec{v} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -t \\ 0 \\ t \end{bmatrix} = t \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}.
$$
A: Youre aproach is right.
The line parallel to both planes or just parallel to the intersection can be found by cross product of normals to planes. ie. $(i+j+k)\times (i-j+k)=(2i-2k)$
Or the direction ratios are$(1,0,1)$ 
So line is $$\vec r=(0+0+0)\pm t(i-k)$$
OR$$x+z=0$$
You see it comes out as a plane $x+0\times y+z=0$ which will pass through line of intersection..See the plot by Wolfram

