= Solution
The right-hand <vector> is in the <column space>, because
$$
2\begin{pmatrix}-1\\1\\2\end{pmatrix}+\begin{pmatrix}3\\2\\5\end{pmatrix}=\begin{pmatrix}1\\4\\9\end{pmatrix}.
$$
Consequently the <linear system> is consistent. A concrete solution is
$$
\boxed{x=(2,0,1,0,0)^{\mathsf T}.}
$$
As a compatibility check, the original rows satisfy $r_1+11r_2-5r_3=0$, and the right-hand entries satisfy the same relation $1+11\cdot4-5\cdot9=0$. There are infinitely many solutions, obtained by adding any <vector> in the <null space> to the displayed solution.
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