Given that T\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x+ 2y- z \\ 2x- y+ z \\ x+ z\end{bmatrix}
Then T\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}= \begin{bmatrix}1+ 2(0)- 1 \\ 2(1)- 0+ 1 \\ 1+ 1\end{bmatrix}= \begin{bmatrix} 0 \\ 3\\ 2 \end{bmatrix}. That is very different from your "\begin{bmatrix}2 \\ 0 \\ 2 \end{bmatrix}". Am I misunderstanding something?
To write that in basis \beta you want to find numbers, A, B, C such that
A\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}+ B\begin{bmatrix}1 \\ 2 \\ 1 \end{bmatrix}+ C\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}= \begin{bmatrix}A+ B+ C \\ 2B+ C \\A+ B \end{bmatrix}= \begin{bmatrix} 0 \\ 3 \\ 2 \end{bmatrix}.
So we have the three equations, A+ B+ C= 0, 2B+ C= 3, and A+ B= 2. From the second equation C= 3- 2B, and from the third, A= 2- B. Putting those into the first equation, A+ B+ C= 2- B+ B+ 3- 2B= 5- 2B= 0. 2B= 5 so B= 2/5. Then C= 3- 4/5= 11/5 and A= 2- 2/5= 8/5. The first column of the transition matrix is \begin{bmatrix}\frac{8}{5} \\ \frac{2}{5} \\ \frac{11}{5}\end{bmatrix}.