My question relates to a simple case of the least squares problem. For example, take the system of equations \begin{align*} 2x- y &= 2, \\ x + 2y &= 1, \\ x+y &= 4, \end{align*} which is clearly overdetermined. Using a least squares approach, we can solve the system $$A^TA\mathbf{x^*} = A^T\mathbf{b},$$ where $\mathbf{x^*}$ minimises $\|\ A\mathbf{x} -\mathbf{b} \ \|$, and $$A = \begin{bmatrix} 2 & -1\\ 1 & 2 \\ 1 & 1 \end{bmatrix} \ , \quad \mathbf{b}= \begin{bmatrix} 2 \\ 1 \\ 4 \end{bmatrix} \quad \mbox{and} \quad \mathbf{x}= \begin{bmatrix} x \\ y \end{bmatrix},$$ to obtain an approximation for the solution as $\left(\frac{10}{7}, \frac{3}{7} \right)$.
If you plot the three straight lines that constitute the system of equations, they intersect at three points that form the vertices of a triangle.
My question is - does the solution always fall within this triangle (as in this example), and if so, is this point a (non-traditional) triangle center?