I just found the least squares solution of the system $\mathbf{x}A = \mathbf{b} = \begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} 3 & 2 & 1 \\ 2 & 3 & 2\end{pmatrix} = \begin{pmatrix} 3 & 0 & 1\end{pmatrix}$ to be $\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} \frac{29}{21} \\ -\frac{2}{3}\end{pmatrix}$. I now am tasked with answering how this solution can be verified without solving for $x$ and $y$.
The system can be rewritten as follows: $\begin{pmatrix}3 & 2 \\ 2 & 3 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix}$.
Since the least-squares solution is the vector $\mathbf{x}$ that makes $\Vert b - A\mathbf{x} \Vert_{2}$ a minimum, should I try graphing the equations $3x + 2y = 3$, $2x + 3y = 0$, and $x + 2y = 1$, then pointing out on the graph that $\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}\frac{29}{21} \\ -\frac{2}{3} \end{pmatrix}$ is close to the point where the three lines intersect?
I figure there must be a more formal way of verifying the solution, but it is escaping me. Could somebody please let me know how I should do it?