How to find the eigenvector when there are multiple instances of the eigenvalues $$\begin{pmatrix}0&1&-1\\2&1&-2\\-1&-1&0\end{pmatrix}$$
The characteristic polynomial is $$\lambda^3-\lambda^2-5\lambda-3=(\lambda+1)(\lambda^2-2\lambda-3)=(\lambda+1)(\lambda+1)(\lambda-3)$$
So I have two eigenvectors with the same eigenvalue $-1$ and I already know one of them is $(1,0,1)^t$.
I found the eigenvector for $3$ but how do I find the other eigenvector for $-1$? If I do it the usual way by placing it in the matrix $(\lambda I -A)$ I get to $$-x-y+z=0$$

When there is an eigenvalue of multiplicity $k>1$, there is an eigenspace of dimension at most $k$. If $k=1$, the dimension is always $1$ as there is always at least one eigenvector $\vec u$, thus all $t\vec u$ are also eigenvectors, thus an eigenspace of dimension $1$.
When $k>1$, it can happen that there are not "enough" eigenvectors (that is, the eigenspace has a dimension $<k$). Here it's not the case though.
You have to solve
$$\left(\begin{matrix}1&1&-1\\2&2&-2\\-1&-1&1\end{matrix}\right)\left(\begin{matrix}x\\y\\z\end{matrix}\right)=\left(\begin{matrix}0\\0\\0\end{matrix}\right)$$
This amount to the single equation $x+y-z=0$, as you found. This means there are two arbitrary parameters, say $x$ and $y$, then $z=x+y$.
To find an eigenspace of dimension $2$, just find two non-collinear eigenvectors, by letting $x=1,y=0$, then $x=0,y=1$, hence the eigenvectors $(1,0,1)^T$ and $(0,1,1)^T$.
Notice there are infinitely many eigenvectors, and any pair of non-collinear vectors lying in the plane defined by the above two would be equally valid.
