Write your system in a matrix form $AX=Y$,
$ \begin{bmatrix}1 & -2\\2 & 1 \\3 & 2 \end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}a \\b \\c\end{bmatrix}\,. $
The technique to find the least square solution of the above system is to apply $A^T$ to both sides of the above equation. Doing that and substituting the given values $x=1$ and $y=-2$ yields
$ \begin{bmatrix}1 & 2 & 3\\-2 & 1 & 2\end{bmatrix}\begin{bmatrix}1 & -2\\2 & 1 \\3 & 2 \end{bmatrix}\begin{bmatrix}1\\-2\end{bmatrix}= \begin{bmatrix}1 & 2 & 3\\-2 & 1 & 2\end{bmatrix}\begin{bmatrix}a \\b \\c\end{bmatrix}\,. $
Work out the above and you will end up having two equations in $a,b,c$ which you need to solve. The solution describes the set of all $(a,b,c)\subset \mathbb{R^3} $