Finding eigenvector only knowing others eigenvectors.

Question

The matrix $A \in M_3(\mathbb{R})$ satisfy $A^t=A$ and $(1,2,1), (-1,1,0)$ are eigenvectors of $A$. Which vector is also an eigenvector of $A$? Alternatives: $(0,0,1)$; $(1,1,-3)$; $(1,1,3)$; There is no other eigenvector.

The problem with this exercise is that I don't know the matrix $A$, and I don't have any eigenvalue to start with. I can get a matrix with less variables using $A = A^t$, but there's still 6 variables. Any tips or guidance is appreciated.

Answer 1

Since $A$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.

So, find which of the vectors is orthogonal to the first two.

(1,1,-3) is.

Answer 2

Hint: the condition $A^t = A$ allows you to use the spectral theorem.

Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $A$.