What are the Eigenvectors of the curl operator?

Be $f(x,y,z)=\mathrm e^{\alpha x+\beta y+\gamma z}$ and $v$ an eigenvector of $\begin{pmatrix} 0 & -\gamma & \beta\\ \gamma & 0 & -\alpha\\ -\beta & \alpha & 0\end{pmatrix}$. Then $f(x,y,z)v$ is an eigenvector of $\rm curl$.

Proof: $\partial_x f(x,y,z) = \alpha f(x,y,z)$, $\partial y f(x,y,z) = \beta f(x,y,z)$, $\partial z f(x,y,z) = \gamma f(x,y,z)$. Therefore the "curl matrix" acts on $f(x,y,z)v$ as if all partial derivatives were replaced by the corresponding factor, which gives the matrix above. Thus, we arrive at an ordinary eigenvector equation, and $v$ (and thus each multiple of $v$) is an eigenvector by assumption.

I'm not sure if those are all eigenvectors, though.