We will appeal to Theorem BS (or you could consider this an appeal to Theorem BCS). Put the three column vectors of this spanning set into a matrix as columns and row-reduce.
\begin{equation*}
\begin{bmatrix}
2 & 3 & -4 \\
-3 & 0 & -3 \\
4 & 1 & 2 \\
1 & -2 & 5
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & 1 \\
0 & \leading{1} & -2 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
\end{equation*}
The pivot columns are $D=\set{1,2}$ so we can “keep” the vectors of $W$ with the same indices and set
\begin{equation*}
T=\set{
\colvector{2\\-3\\4\\1},\,
\colvector{3\\0\\1\\-2}}
\end{equation*}
and conclude that $W=\spn{T}$ and $T$ is linearly independent. In other words, $T$ is a basis with two vectors, so $W$ has dimension 2.