Form a linear combination, with unknown scalars, of $S$ that equals $\vect{w}$,
\begin{equation*}
a_1\colvector{-1 \\ 2 \\ 1}+
a_2\colvector{ 3 \\ 1 \\ 2}+
a_3\colvector{ 1 \\ 5 \\ 4}+
a_4\colvector{-6 \\ 5 \\ 1}
=
\colvector{2\\1\\3}
\end{equation*}
We want to know if there are values for the scalars that make the vector equation true since that is the definition of membership in $\spn{S}$. By Theorem SLSLC any such values will also be solutions to the linear system represented by the augmented matrix,
\begin{equation*}
\begin{bmatrix}
-1 & 3 & 1 & -6 & 2 \\
2 & 1 & 5 & 5 & 1 \\
1 & 2 & 4 & 1 & 3
\end{bmatrix}
\end{equation*}
Row-reducing the matrix yields,
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 2 & 3 & 0 \\
0 & \leading{1} & 1 & -1 & 0 \\
0 & 0 & 0 & 0 & \leading{1}
\end{bmatrix}
\end{equation*}
With a pivot column in the last column, the system is inconsistent (Theorem RCLS), so there are no scalars $a_1,\,a_2,\,a_3,\,a_4$ that will create a linear combination of the vectors in $S$ that equal $\vect{w}$. So $\vect{w}\not\in\spn{S}$.