Find a any matrix $C$ satisfying the below conditions. Let the $A,B$ and $C$ are $3\times 3$ matrices.
For the vectors $v_1= \begin{bmatrix} 1 \\ 1 \\ -1 \\ \end{bmatrix}$,$v_2= \begin{bmatrix} 0 \\ 1 \\ -1 \\ \end{bmatrix}$ and $v_3= \begin{bmatrix} 1 \\ 1 \\ -2 \\ \end{bmatrix}$, $Av_1=v_1 
Av_2=v_2, Av_3=-2v_3, Bv_1 = -v_1, Bv_2 = 2v_2$ and $Bv_3 = -v_3$. Find a any $C$ whose eigenvalues are $0,1,-1$ with $AC=CA$ and $BC=CB$.
I take the case $v_1, v_2$ and $v_3$ are eigenvectors of the $0,1$ and $-1$ respectively.
So Putting $P= [v_1 \vert v_2 \vert v_3 ]= \begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ -1 & -1& -2 \\\end{bmatrix}$ and $D =  \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0& -1 \\\end{bmatrix}$.
From those we get $P^{-1} = \begin{bmatrix} 1 & 1 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & -1 \\\end{bmatrix}$. If we take the $C = P^{-1}DP$ the conditions of the $AC=CA$ and $BC=CB$ would satisfy because $A, B$ are diagonalizable considering the $P$ (e.g $PD_1P^{-1} =A$)
Consequently my final answer is  $C = P^{-1}DP = \begin{bmatrix} -1 & 1 & 1 \\ -2 & 2 & 1 \\ 3 & -3 & -2 \\\end{bmatrix}$. But the eigenvalues of the $C$ which I made are  $0,1$ and $2$. I can't find any error in my solution. Please let me know my mistake.

$$C = P^{-1}DP=\left(\begin{bmatrix} 1 & 1 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & -1 \\\end{bmatrix}\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0& -1 \\\end{bmatrix}\right)\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ -1 & -1& -2 \\\end{bmatrix}$$
$$=\left(\begin{bmatrix} 0 & 1 & -1 \\ 0 & 1 & 0\\ 0& -1& 1 \\\end{bmatrix}\right)\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ -1 & -1& -2 \\\end{bmatrix}$$
$$= \begin{bmatrix} 2 & 2 & 3 \\ 1 & 1 & 1 \\ -2 & -2& -3 \\\end{bmatrix}.$$
