$\left[\begin{array}{ccc}1 & 2 & -1 \\ 3 & -1 & 2 \\ 2 & -2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}3 \\ 1 \\ 2\end{array}\right]$
Augmented matrix is:
$\begin{aligned}
& {\left[\begin{array}{ccccc}
1 & 2 & -1 & : & 3 \\
3 & -1 & 2 & : & 1 \\
2 & -2 & 3 & : & 2
\end{array}\right]} \\
& R_2 \rightarrow R_2-3 R_1 \text { \& } R_3 \rightarrow R_3-2 R_1 \\
& \sim\left[\begin{array}{ccccc}
1 & 2 & -1 & : & 3 \\
0 & -7 & 5 & : & -8 \\
0 & -6 & 5 & : & -4
\end{array}\right] \\
& R_3 \rightarrow 7 R_3-6 R_2 \\
& \sim\left[\begin{array}{ccccc}
1 & 2 & -1 & : & 3 \\
0 & -7 & 5 & : & -8 \\
0 & 0 & 5 & : & 20
\end{array}\right]
\end{aligned}$
$\because \operatorname{Rank}(A: B)=\operatorname{Rank}(A)=3$
$\therefore \quad$ Solution exists and can be given by :
$\begin{aligned}
& \therefore\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & -7 & 5 \\
0 & 0 & 5
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
3 \\
-8 \\
20
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{l}
x+2 y-z \\
0-7 y+5 z \\
0+0+5 z
\end{array}\right]=\left[\begin{array}{c}
3 \\
-8 \\
20
\end{array}\right] \\
& \therefore \quad x+2 y-z=3 \\
& -7 y+5 z=-8 \\
& 5 z=20 \Rightarrow z=4 \\
& \therefore \quad-7 y+20=-8 \Rightarrow y=4 \\
& x+8-4=3 \Rightarrow x=1 \\
& \therefore \alpha=1, \beta=4, \gamma=4 \\
& \therefore \quad \alpha^2+\beta^2+\gamma^2=1+16+16=33 \\
&
\end{aligned}$