Answer
$(x,y,z)=(4,-3,3)$
Work Step by Step
Formula to determine the determinant, $D$ of a $3 \times 3$ matrix is:
$D=\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=a \begin{vmatrix}e&f\\h&i\end{vmatrix}-b \begin{vmatrix}d&f\\g&i\end{vmatrix}+c \begin{vmatrix}d&e\\g&h\end{vmatrix}$
Need to apply Cramer's Rule.
$x=\dfrac{D_x}{D};y=\dfrac{D_y}{D}; z=\dfrac{D_z}{D}$
Now
$D=\begin{vmatrix}2&3&1\\3&3&-1\\1&-2&-3\end{vmatrix}=-7$;
and $D_x=\begin{vmatrix}2&3&1\\0&3&-1\\1&-2&-3\end{vmatrix}=-28$;
$D_y=\begin{vmatrix}2&2&1\\3&0&-1\\1&1&-3\end{vmatrix}=21$
$D_z=\begin{vmatrix}2&3&2\\3&3&0\\1&-2&1\end{vmatrix}=-21$
Thus,
$x=\dfrac{-28}{-7}=4;y=\dfrac{21}{-7}=-3; z=\dfrac{-21}{-7}=3$
Hence, $(x,y,z)=(4,-3,3)$
