Since $3 x+2 y-2 z=3$
$x+2 y+3 z=6$
and $2 x-y+z=2$
In the form of $\mathrm{AX}=\mathrm{B}$,
$\left[\begin{array}{ccc}3 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & -1 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\left[\begin{array}{l}3 \\ 6 \\ 2\end{array}\right]$
For $\mathrm{A}^{-1},|\mathrm{~A}|=|3(5)-2(1-6)+(-2)(-5)|=|35| \neq 0$
$\therefore \mathrm{A}_{11}=5, \mathrm{~A}_{12}=5, \mathrm{~A}_{13}=-5, \mathrm{~A}_{21}=0, \mathrm{~A}_{22}, 7, \mathrm{~A}_{23}=7, \mathrm{~A}_{31}$
$=10, \mathrm{~A}_{32}=-11$ and $\mathrm{A}_{33}=4$
$\therefore \quad \operatorname{adj} A=\left[\begin{array}{ccc}5 & 5 & -5 \\ 0 & 7 & 7 \\ 10 & -11 & 4\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4\end{array}\right]$
Now, $\quad \mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}=\frac{1}{35}\left[\begin{array}{ccc}5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4\end{array}\right]$
For $\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}$.


$$
\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\frac{1}{35}\left[\begin{array}{ccc}
5 & 0 & 10 \\
5 & 7 & -11 \\
-5 & 7 & 4
\end{array}\right]\left[\begin{array}{l}
3 \\
6 \\
2
\end{array}\right]=\frac{1}{35}\left[\begin{array}{l}
35 \\
35 \\
35
\end{array}\right]=\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]
$$
$\therefore \mathrm{x}=1, \mathrm{y}=1$ and $\mathrm{z}=1$