Theorem BRS is the most direct route to a set with these properties. Row-reduce, toss zero rows, keep the others. You could also transpose the matrix, then look for the column space by row-reducing the transpose and applying Theorem BCS. We will do the former,
\begin{equation*}
B\rref
\begin{bmatrix}
\leading{1} & 0 & -1 & 2\\
0 & \leading{1} & 1 & -1\\
0 & 0 & 0 & 0
\end{bmatrix}
\end{equation*}
So the set $S$ is
\begin{equation*}
S=\set{
\colvector{1\\0\\-1\\2},\,
\colvector{0\\1\\1\\-1}
}
\end{equation*}