$\ker(T) = \{(x,y,z): T(x,y,z) = (0,0,0)\} = \{(x,y,z): x=0, y=0\} = \{(0,0,z): z\in\mathbb{R}\}$. This is the $z$-axis — a line through the origin. Wait — the $z$-axis is a line. But the answer key says plane. Rechecking: $T(x,y,z)=(x,y,0)=\mathbf{0}$ requires $x=0, y=0$; $z$ is free. This is a line (the $z$-axis).