In practice, one uses the singular value decomposition, $\mathbf A=\mathbf U\mathbf \Sigma\mathbf V^\top$ for solving underdetermined problems like these. Taking the SVD approach assumes that you are optimizing with respect to the Euclidean norm, $\|\cdot\|_2$. (If you need to optimize with respect to the 1-norm or max-norm, linear programming methods are required, but I won't get into those.)
For this particular example, we have the decomposition
$\begin{align*} \mathbf U&=\begin{pmatrix} 0.4964775289157638 & -0.8680495742074279 \\ 0.8680495742074279 & 0.4964775289157638 \end{pmatrix} \\ \mathbf \Sigma&=\begin{pmatrix} 6.302625081925469 & 0 \\ 0 & 0.5262291104490325 \end{pmatrix} \\ \mathbf V&=\begin{pmatrix} 0.2165013919416455 & -0.7061031742896186 \\ 0.3542296500759905 & 0.2373595096582885 \\ 0.5707310420176360 & -0.4687436646313301 \\ 0.7084593001519810 & 0.4747190193165770 \end{pmatrix} \end{align*} $
Computing the least-squares solution $\min\|\mathbf A\mathbf x-\mathbf b\|_2$ is a matter of computing $\mathbf x=\mathbf V\mathbf \Sigma^{-1}\mathbf U^\top\mathbf b$; for the particular case of $\mathbf b=(0\quad 11)^\top$, we obtain the solution $\mathbf x=\pmatrix{-7\\3\\-4\\6}$ There are other solutions, like $\mathbf x=(0\quad 0\quad -11\quad 11)^\top$. All take the form $\mathbf x=\left(a\quad b\quad -11-a\quad 11+\frac{a-b}{2}\right)^\top$