Seriously? Your matrix is the 0 matrix. Is it not obvious that the 0 matrix times any vector is 0? Yes, \(\displaystyle <x_1, x_2, 0>\) is an eigenvector for any \(\displaystyle x_1\) or \(\displaystyle x_2\) but that is because any vector, of the form \(\displaystyle <x_1, x_2, x_3>\), for any \(\displaystyle x_1\), \(\displaystyle x_2\), \(\displaystyle x_3\), is an eigenvector!
It looks to me like you have fallen into the trap of memorizing formulas rather than the learning the concepts. You should have been able to look at the 0 matrix and, from the basic definitions of "eigenvalues" and "eigenvectors" have realized that "0" is the only eigenvalue and that all vectors are eigenvectors.
(Most text books define an eigenvalue by "\(\displaystyle \lambda\) is an eigenvalue of A if there exist a non-zero vector v, such that \(\displaystyle Av= \lambda v\)" and then define an eigenvector, corresponding to eigenvalue \(\displaystyle \lambda\), as any such non-zero vector. I, and a few textbooks prefer to include the "non-zero" part in the definition of "eigenvalue" but not in the definition of "eigenvalue". That way we can say that "the set of all eigenvectors corresponding to a given eigenvalue form a subspace" rather than having to say that "the set of all eigenvectors corresponding to a given eigenvalue, together with the 0 vector, form a subspace". If you use the more common definition of "eigenvector", you would have to say that "any vector except the 0 vector is an eigenvector for the 0 matrix".)