Rephrasing the question, we want to know if there are scalars $\alpha_1$ and $\alpha_2$ such that
\begin{equation*}
\alpha_1\colvector{2\\-1\\3\\4}+\alpha_2\colvector{3\\2\\-2\\1}=\colvector{5\\1\\3\\5}
\end{equation*}
Theorem SLSLC allows us to rephrase the question again as a quest for solutions to the system of four equations in two unknowns with an augmented matrix given by
\begin{equation*}
\begin{bmatrix}
2 & 3 & 5\\
-1 & 2 & 1\\
3 & -2 & 3\\
4 & 1 & 5
\end{bmatrix}
\end{equation*}
This matrix row-reduces to
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 0\\
0 & \leading{1} & 0\\
0 & 0 & \leading{1}\\
0 & 0 & 0
\end{bmatrix}
\end{equation*}
With a pivot column in the last column of this matrix (Theorem RCLS) we can see that the system of equations has no solution, so there are no values for $\alpha_1$ and $\alpha_2$ that will allow us to conclude that $\vect{y}$ is in $W$. So $\vect{y}\not\in W$.