What do you mean "the answer is ..."? Do you understand what an eigenvector is? If a given vector is an eigenvector so is any multiple of it. You cannot just say "the eigenvector" is any specific vector. I said before, "we must have z= 0. We have NO information about x or y so they can be anything."
That includes your [a, 0, 0], taking x= a, y= 0. It also includes [1, 0, 0], [0, 0 1], and any linear combination x[1, 0, 0]+ y[0, 1, 0]= [x, y, 0].
For any x, y,
\begin{bmatrix}0 & 0 & a \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x \\ y \\ 0\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}= 0\begin{bmatrix}x \\ y\\ 0\end{bmatrix}