 Solution

Minimize $\frac{1}{2}\Vert \mathbf{r}{\Vert }^2$ subject to ${\mathbf{n}}_1\cdot \mathbf{r}=1$, ${\mathbf{n}}_2\cdot \mathbf{r}=2$, where ${\mathbf{n}}_1=(2,1,-1)$ and ${\mathbf{n}}_2=(1,-1,1)$. The Lagrange multiplier equation gives $\mathbf{r}=\lambda {\mathbf{n}}_1+\mu {\mathbf{n}}_2$. Since ${\mathbf{n}}_1\cdot {\mathbf{n}}_2=0$, $\Vert {\mathbf{n}}_1{\Vert }^2=6$, and $\Vert {\mathbf{n}}_2{\Vert }^2=3$, the constraints imply $\lambda =1/6$, $\mu =2/3$. HenceNo explicit parametrization of the line was needed. For any other feasible point, $\mathbf{h}=\mathbf{r}-{\mathbf{r}}_{\ast }$ is orthogonal to both normals and therefore to ${\mathbf{r}}_{\ast }$. The Pythagorean theorem gives $\Vert \mathbf{r}{\Vert }^2=\Vert {\mathbf{r}}_{\ast }{\Vert }^2+\Vert \mathbf{h}{\Vert }^2$, proving the unique global minimum. This is the geometric principle behind the minimum-norm least-squares solution.