Answer
$a.\quad $
A plane, perpendicular to the yz plane, intersecting the plane yz in the line $z=1-y.$
$ b.\quad$
The cubic curve $z=y^{3}$ in the plane $x=2$ (which is parallel to the yz plane).
Work Step by Step
$a.\quad $
In the plane $x=0$ (the yz plane),$ z=1-y $ is a line.
Translating this line across planes $x=k$ (parallel to the yz plane), a plane is formed, perpendicular to the yz plane, intersecting the plane yz in the line $z=1-y.$
$ b.\quad$
In the plane $x=0$ the equation $z=y^{3}$ is a cubic curve.
Translating along the x axis by 2 units, this curve is now in the plane $x=2.$
