Theorem BS says that if we take these 5 vectors, put them into a matrix, and row-reduce to discover the pivot columns, then the vectors of $S$ with the same column indices will be linearly independent and span $S$, and thus will form a basis of $S$.
\begin{align*}
\begin{bmatrix}
1 & 1 & 1 & 1 & 3\\
3 & 2 & 1 & 2 & 4\\
2 & 1 & 0 & 2 & 1\\
1 & 1 & 1 & 1 & 3
\end{bmatrix}
&\rref
\begin{bmatrix}
\leading{1} & 0 & -1 & 0 & -2\\
0 & \leading{1} & 2 & 0 & 5\\
0 & 0 & 0 & \leading{1} & 0\\
0 & 0 & 0 & 0 &0
\end{bmatrix}
\end{align*}
Since columns 1, 2, and 4 are pivot columns, the vectors that span $S$ are the first, second and fourth of the set, so a basis of $S$ is
\begin{align*}
B &= \set{
\colvector{1\\3\\2\\1},
\colvector{1\\2\\1\\1},
\colvector{1\\2\\2\\1}
}
\end{align*}