Answer
Solution set = $\{( 2 ,\ -1 ,\ 1 )\}$
Work Step by Step
Reduce the augmented matrix $[A|B]$ to reduced row echelon form and interpret the result
$\left[\begin{array}{lllll}
1 & -2 & 3 & | & 7\\
2 & 1 & 1 & | & 4\\
-3 & 2 & -2 & | & -10
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{2}=r_{2}-2r_{1}.\\
R_{3}=r_{3}+3r_{1}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & -2 & 3 & | & 7\\
0 & 5 & -5 & | & -10\\
0 & -4 & 7 & | & 11
\end{array}\right]\rightarrow\left(\begin{array}{l}
\\
R_{2}=\frac{1}{5}r_{2}.\\
.
\end{array}\right) $
$\rightarrow\left[\begin{array}{lllll}
1 & -2 & 3 & | & 7\\
0 & 1 & -1 & | & -2\\
0 & -4 & 7 & | & 11
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}=r_{1}+2r_{2}.\\
.\\
R_{3}=r_{3}+4r_{2}.
\end{array}\right) $
$\rightarrow\left[\begin{array}{lllll}
1 & 0 & 1 & | & 3\\
0 & 1 & -1 & | & -2\\
0 & 0 & 3 & | & 3
\end{array}\right] \rightarrow\left(\begin{array}{l}
.\\
.\\
R_{3}=\frac{1}{3}r_{3}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & 0 & 1 & | & 3\\
0 & 1 & -1 & | & -2\\
0 & 0 & 1 & | & 1
\end{array}\right] \rightarrow\left(\begin{array}{l}
R_{1}=r_{1}-r_{3}.\\
R_{2}=r_{2}+r_{3}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & 0 & 0 & | & 2\\
0 & 1 & 0 & | & -1\\
0 & 0 & 1 & | & 1
\end{array}\right]$
The system is consistent and has a single solution.
$x=2,$
$y=-1$
$z=1$
Solution set = $\{( 2 ,\ -1 ,\ 1 )\}$
