The conventional thing to do is to expand $\det(A-\lambda I)$ and factor it. For artificial exercises like this one, the roots are often “nice,” so you can use the rational root theorem to make some educated guesses as to what the roots might be.
Before diving into all of that, it can be fruitful to try to find some eigenvectors by playing around with simple linear combinations or columns and rows of the matrix. The eigenvalues of a matrix and its transpose are identical, so left eigenvectors—eigenvectors of the transpose—are just as good if we’re mainly interested in the eigenvalues.
Here, we can quickly find that adding the first and second rows produces $(2,2,0)$, so $(1,1,0)$ is a left eigenvector with eigenvalue $2$. Similarly, adding the first and third rows produces $(-2,0,-2)$, so $(1,0,1)$ is a left eigenvector with eigenvalue $-2$. You can get the last eigenvalue “for free” by using the fact that the trace of a matrix is equal to the sum of its eigenvalues: $(1+3-1)-2+2=3$.
Therefore, the characteristic polynomial is $(\lambda+2)(\lambda-2)(\lambda-3)$.