Solution:
Given that $3 x-y+2 z=0$
$\begin{array}{l}
4 x+3 y+3 z=0 \\
5 x+7 y+4 z=0
\end{array}$
We can write the system as
$\begin{array}{l}
{\left[\begin{array}{ccc}
3 & -1 & 2 \\
4 & 3 & 3 \\
5 & 7 & 4
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]} \\
\mathrm{A} X=0
\end{array}$
Now, $|A|=3(12-21)+1(16-15)+2(28-15)$
$\begin{array}{l}
|\mathrm{A}|=-27+1+26 \\
|\mathrm{~A}|=0
\end{array}$
As a result, the system has infinite solutions
Suppose $z=k$
$\begin{array}{l}
3 x-y=-2 k \\
4 x+3 y=-3 k
\end{array}$
$\left[\begin{array}{cc}
3 & -1 \\
4 & 3
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y}
\end{array}\right]=\left[\begin{array}{l}
-2 \mathrm{k} \\
-3 \mathrm{k}
\end{array}\right]$
$\begin{array}{l}
A X=B \\
|A|=9+4=13 \neq 0 \text { So, } A^{-1} \text { exist }
\end{array}$
Now adj $A=\left[\begin{array}{cc}3 & -1 \\ 4 & 3\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}3 & 1 \\ -4 & 3\end{array}\right]$
$\begin{array}{l}
X=A^{-1} B=\frac{1}{|A|}(\operatorname{adj} A) B=\frac{1}{13}\left[\begin{array}{cc}
3 & 1 \\
-4 & 3
\end{array}\right]\left[\begin{array}{l}
-2 k \\
-3 k
\end{array}\right] \\
X=\left[\frac{-9 k}{13}\right]
\end{array}$
As a result, $x=\frac{-9 \mathrm{k}}{13}, \mathrm{y}=\frac{-\mathrm{k}}{13}$ and $\mathrm{z}=\mathrm{k}$