Solution:
Given that $x+y-6 z=0$
$\begin{array}{l}
x-y+2 z=0 \\
-3 x+y+2 z=0
\end{array}$
We can write the system as
$\begin{array}{l}
{\left[\begin{array}{ccc}
1 & 1 & -6 \\
1 & -1 & 2 \\
-3 & 1 & 2
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]} \\
\mathrm{A} \times=0 \\
\text { Now, }|\mathrm{A}|=1(-2-2)-1(2+6)-6(1-3) \\
|\mathrm{A}|=-4-8+12 \\
|\mathrm{~A}|=0
\end{array}$
As a result, the system has infinite solutions
Suppose $z=k$
$\begin{array}{l}
x+y=6 k \\
x-y=-2 k \\
{\left[\begin{array}{cc}
1 & 1 \\
1 & -1
\end{array}\right]\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{y}
\end{array}\right]=\left[\begin{array}{c}
6 \mathrm{k} \\
-2 \mathrm{k}
\end{array}\right]} \\
\mathrm{A} \mathrm{X}=\mathrm{B} \\
|\mathrm{A}|=-1-1=-2 \neq 0 \mathrm{So}, \mathrm{A}^{-1} \text { exist }
\end{array}$
Now adj $A=\left[\begin{array}{cc}-1 & -1 \\ -1 & 1\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}-1 & -1 \\ -1 & 1\end{array}\right]$
$X=A^{-1} B=\frac{1}{|A|}(\operatorname{adj} A) B=\frac{1}{-2}\left[\begin{array}{cc}
-1 & -1 \\
-1 & 1
\end{array}\right]\left[\begin{array}{c}
6 \mathrm{k} \\
-2 \mathrm{k}
\end{array}\right]$
$\mathrm{X}=\frac{1}{-2}\left[\begin{array}{l}
-6 \mathrm{k}+2 \mathrm{k} \\
-6 \mathrm{k}-2 \mathrm{k}
\end{array}\right]$
$X=\left[\begin{array}{l}
-4 k \\
-8 k
\end{array}\right]$
As a result, $x=2 k, y=4 k$ and $z=k$