Answer
$ x=-1 , y=3, z=2$
Work Step by Step
First we find $D, D_x , D_y $ and $D_z$.
$$D=\begin{vmatrix}
1 & 2 & -1\\
2 & -3 & 1\\
-1&1&-2
\end{vmatrix}=1\begin{vmatrix}
-3& 1 \\
1 & -2
\end{vmatrix}-2\begin{vmatrix}
2& 1 \\
-1 & -2
\end{vmatrix}+(-1)\begin{vmatrix}
2& -3 \\
-1 & 1
\end{vmatrix}$$
$$=1(6-1)-2(-4+1)-1(2-3)=5+6+1=12$$
$$D_x=\begin{vmatrix}
3 & 2 & -1\\
-9 & -3 & 1\\
0&1&-2
\end{vmatrix}=3\begin{vmatrix}
-3& 1 \\
1 & -2
\end{vmatrix}-2\begin{vmatrix}
-9& 1 \\
0 & -2
\end{vmatrix}+(-1)\begin{vmatrix}
-9& -3 \\
0 & 1
\end{vmatrix}$$
$$=3(6-1)-2(18-0)-1(-9-0)=15-36+9=-12$$
$$D_y=\begin{vmatrix}
1 & 3 & -1\\
2 & -9 & 1\\
-1&0&-2
\end{vmatrix}=1\begin{vmatrix}
-9& 1 \\
0 & -2
\end{vmatrix}-3\begin{vmatrix}
2& 1 \\
-1 & -2
\end{vmatrix}+(-1)\begin{vmatrix}
2& -9 \\
-1 & 0
\end{vmatrix}$$
$$=1(18-0)-3(-4+1)-1(0-9)=18+9+9=36$$
$$D_z=\begin{vmatrix}
1 & 2 & 3\\
2 & -3 & -9\\
-1&1&0
\end{vmatrix}=1\begin{vmatrix}
-3& -9 \\
1 & 0
\end{vmatrix}-2\begin{vmatrix}
2& -9 \\
-1 & 0
\end{vmatrix}+3\begin{vmatrix}
2& -3 \\
-1 & 1
\end{vmatrix}$$
$$=1(0+9)-2(0-9)+3(2-3)=9+18-3=24$$
Now,
$$x=\frac{D_x}{D}=\frac{-12}{12}=-1$$
$$y=\frac{D_y}{D}=\frac{36}{12}=3$$
$$z=\frac{D_z}{D}=\frac{24}{12}=2$$
