## Calculus (3rd Edition)

Not parallel: (c) $x-y+z=0$.
(c) $x-y+z=0$. Since the normal vector to the given plane is $\langle 1,1,1 \rangle$, then for any parallel plane its normal vector must be in the direction of $\langle 1,1,1 \rangle$. However, the plane in (c) has the normal vector $\langle 1,-1,1 \rangle$, which is not in the direction of $\langle 1,1,1 \rangle$.