Form a linear combination, with unknown scalars, of $R$ that equals $\vect{z}$,
\begin{equation*}
a_1\colvector{2\\-1\\3\\4\\0}+
a_2\colvector{1\\1\\2\\2\\-1}+
a_3\colvector{3\\-1\\0\\3\\-2}
=
\colvector{1\\1\\5\\3\\1}
\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 $\spn{R}$. By Theorem SLSLC any such values will also be solutions to the linear system represented by the augmented matrix,
\begin{equation*}
\begin{bmatrix}
2 & 1 & 3 & 1\\
-1 & 1 & -1 & 1\\
3 & 2 & 0 & 5\\
4 & 2 & 3 & 3\\
0 & -1 & -2 & 1
\end{bmatrix}
\end{equation*}
Row-reducing the matrix 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*}
With a pivot column in the last column, the system is inconsistent (Theorem RCLS), so there are no scalars $a_1,\,a_2,\,a_3$ that will create a linear combination of the vectors in $R$ that equal $\vect{z}$. So $\vect{z}\not\in R$.