Theorem BCS is the right tool for this problem. Row-reduce this matrix, identify the pivot columns and then grab the columns of $A$ with the same indices for the set $T$. The matrix $A$ row-reduces to
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 3 & 0 & 0 \\
0 & \leading{1} & -2 & 0 & 0 \\
0 & 0 & 0 & \leading{1} & 0 \\
0 & 0 & 0 & 0 & \leading{1}
\end{bmatrix}
\end{equation*}
So $D=\set{1,\,2,\,4,\,5}$ and then
\begin{equation*}
T=\set{\vect{A}_1,\,\vect{A}_2,\,\vect{A}_4,\,\vect{A}_5}=
\set{
\colvector{2 \\ 1 \\ -1 \\ 2},\,
\colvector{1 \\ -1 \\ 2 \\ -1},\,
\colvector{-1 \\ 1 \\ 0 \\ -1},\,
\colvector{2 \\ 1 \\ 1 \\ 2}
}
\end{equation*}
has the requested properties.