Theorem BS says we can make a matrix with these four vectors as columns, row-reduce, and just keep the columns with indices in the set $D$. Here we go, forming the relevant matrix and row-reducing,
\begin{equation*}
\begin{bmatrix}
2 & 3 & 1 & 5 \\
-1 & 0 & 1 & -1 \\
2 & 1 & -1 & 3
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & -1 & 1 \\
0 & \leading{1} & 1 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}
\end{equation*}
Analyzing the row-reduced version of this matrix, we see that the first two columns are pivot columns, so $D=\set{1,2}$. Theorem BS says we need only “keep” the first two columns to create a set with the requisite properties,
\begin{equation*}
T=\set{
\colvector{2\\-1\\2},\,
\colvector{3\\0\\1}
}
\end{equation*}