Solution:

It is given that
$\begin{array}{l}
2 x-y+z=0 \\
3 x+2 y-z=0 \\
x+4 y+3 z=0
\end{array}$
We can write the system as
$\left[\begin{array}{ccc}
2 & -1 & 1 \\
3 & 2 & -1 \\
1 & 4 & 3
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{Z}
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]$
$A X=0$
Now, $|\mathrm{A}|=2(6+4)+1(9+1)+1(12-2)$
$\begin{array}{l}
|A|=2(10)+10+10 \\
|A|=40 \neq 0
\end{array}$
As, $|\mathrm{A}| \neq 0$, as a result $\mathrm{x}=\mathrm{y}=\mathrm{z}=0$ is the only solution of this homogeneous equation.