Theorem LIVRN suggests we analyze a matrix whose columns are the vectors of $S$,
\begin{equation*}
A=\begin{bmatrix}
1 & 3 & 2 & 1\\
-2 & 3 & 1 & 0\\
2 & 1 & 2 & 1\\
5 & 2 & -1 & 2\\
3 & -4 & 1 & 2
\end{bmatrix}
\end{equation*}
Row-reducing the matrix $A$ yields,
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 0 & 0\\
0 & \leading{1} & 0 & 0\\
0 & 0 & \leading{1} & 0\\
0 & 0 & 0 & \leading{1}\\
0 & 0 & 0 & 0
\end{bmatrix}
\end{equation*}
We see that $r=4=n$, where $r$ is the number of nonzero rows and $n$ is the number of columns. By Theorem LIVRN, the set $S$ is linearly independent.