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\\8\\-12\\-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 & 8\\
3 & -2 & -12\\
4 & 1 & -5
\end{bmatrix}
\end{equation*}
This matrix row-reduces to
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & -2\\
0 & \leading{1} & 3\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\end{equation*}
From the form of this matrix, we can see that $\alpha_1=-2$ and $\alpha_2=3$ is an affirmative answer to our question. More convincingly,
\begin{equation*}
(-2)\colvector{2\\-1\\3\\4}+(3)\colvector{3\\2\\-2\\1}=\colvector{5\\8\\-12\\-5}
\end{equation*}