I do not think your approach is correct.
We can simplify the answer by taking the ordered bases
$$\begin{align}
\mathscr B_3' & = \left\{
\varepsilon_1 = \begin{pmatrix}1\\0\\0\end{pmatrix},
\varepsilon_2 = \begin{pmatrix}0\\1\\0\end{pmatrix},
\varepsilon_3 = \begin{pmatrix}0\\0\\1\end{pmatrix}\right\},\\\\
\mathscr B_2' & = \left\{
\begin{pmatrix}\beta_{11}\\ \beta_{12} \end{pmatrix},
\begin{pmatrix}\beta_{21}\\ \beta_{22} \end{pmatrix}\right\}\end{align}$$
and solving for $\mathscr B_2'.$ The simplification is that with our choice for $\mathscr B_3',$ coordinate matrices in $\Bbb R^3$ have the same entries as vectors in $\Bbb R^3,$ so we do not have to worry about converting from one to the other.
Now the $i$th column of Mat$_{\mathscr B_3', \mathscr B_2'}(T)$ is $[T\varepsilon_i]_{\mathscr B_2'}$ where $[ \cdot ]_{\mathscr B_2'}$ denotes the coordinate matrix with respect to ordered basis $\mathscr B_2'.$ That gives us the following three systems of equations:
$$\begin{align}
\begin{pmatrix} 1\\ 0\end{pmatrix} & = \left[ T \begin{pmatrix} 1\\ 0\\ 0\end{pmatrix} \right]_{\mathscr B_2'}\\\\
& = \left[ \begin{pmatrix} 2\\ 0\end{pmatrix} \right]_{\mathscr B_2'}\\\\
1 \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} +
0 \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix} & =
\begin{pmatrix} 2 \\ 0 \end{pmatrix},\\\\
\end{align}$$
$$\begin{align}
\begin{pmatrix} 0\\ 1\end{pmatrix} & = \left[ T \begin{pmatrix} 0\\ 1\\ 0\end{pmatrix} \right]_{\mathscr B_2'}\\\\
& = \left[ \begin{pmatrix} 1\\ 1\end{pmatrix} \right]_{\mathscr B_2'}\\\\
0 \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} +
1 \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix} & =
\begin{pmatrix} 1 \\ 1 \end{pmatrix},\\\\
\end{align}$$
$$\begin{align}
\begin{pmatrix} -1\\ 2\end{pmatrix} & = \left[ T \begin{pmatrix} 0\\ 0\\ 1\end{pmatrix} \right]_{\mathscr B_2'}\\\\
& = \left[ \begin{pmatrix} 0\\ 2\end{pmatrix} \right]_{\mathscr B_2'}\\\\
-1 \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} +
2 \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix} & =
\begin{pmatrix} 0 \\ 2 \end{pmatrix}.\end{align}$$
Solving for the four $\beta$s, we get
$$\mathscr B_2' = \left\{
\begin{pmatrix} 2\\ 0\end{pmatrix},
\begin{pmatrix} 1\\ 1\end{pmatrix}\right\}.$$
As amd stated in his answer to your previous question, other bases will satisfy the hypotheses of your question.