We have AX = B
Where, $A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$
and $B=\left[\begin{array}{c}11 \\ -5 \\ -3\end{array}\right]$
$\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \quad \ldots(i)$
Now, $|\mathrm{A}|=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]=-1 \neq 0$
$\therefore \mathrm{A}^{-1}$ non-cofactory case and hence the given equations have a unique solution.
$\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A})=\frac{1}{-1}\left[\begin{array}{ccc}0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13\end{array}\right]$
$=\frac{1}{-1}\left[\begin{array}{ccc}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]$
From(i)
$\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right]\left[\begin{array}{c}
11 \\
-5 \\
-3
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]$
$\Rightarrow x=1, y=2$ and $z=3$