$3 x-y+2 z=0$
$4 x+3 y+3 z=0$
$5 x+7 y+4 z=0$
Here,
$3 x-y+2 z=0$ ....(1)
$4 x+3 y+3 z=0$ .....(2)
$5 x+7 y+4 z=0$ ....(3)
The given system of homogeneous equations can be written in matrix form as follows:
$\left[\begin{array}{ccc}3 & -1 & 2 \\ 4 & 3 & 3 \\ 5 & 7 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
$A X=O$
Here,
$A=\left[\begin{array}{ccc}3 & -1 & 2 \\ 4 & 3 & 3 \\ 5 & 7 & 4\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $O=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
Now,
$|A|=\left|\begin{array}{ccc}3 & -1 & 2 \\ 4 & 3 & 3 \\ 5 & 7 & 4\end{array}\right|$
$=3(12-21)+1(16-15)+2(28-15)$
$=-27+1+26$
$=0$
$\therefore|A| \neq 0$
So, the given system of homogeneous equations has non-trivial solution.
Substituting $z=k$ in eq. (1) \& eq. (2), we get
$3 x-y=-2 k$ and $4 x+3 y=-3 k$
$A X=B$
Here,
$\mathrm{A}=\left[\begin{array}{cc}3 & -1 \\ 4 & 3\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}-2 k \\ -3 k\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}3 & -1 \\ 4 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}-2 k \\ -3 k\end{array}\right]$
$|A|=\left|\begin{array}{cc}3 & -1 \\ 4 & 3\end{array}\right|$
$=(3 \times 3+4 \times 1)$
$=13$
So, $A^{-1}$ exists.
We have
$\operatorname{adj} A=\left[\begin{array}{cc}3 & 1 \\ -4 & 3\end{array}\right]$
$A^{-1}=\frac{1}{|A|}$ adj $A$
$\Rightarrow A^{-1}=\frac{1}{13}\left[\begin{array}{cc}3 & 1 \\ -4 & 3\end{array}\right]$
$X=A^{-1} B$
$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{13}\left[\begin{array}{cc}3 & 1 \\ -4 & 3\end{array}\right]\left[\begin{array}{l}-2 k \\ -3 k\end{array}\right]$
$=\frac{1}{13}\left[\begin{array}{c}-6 k-3 k \\ 8 k-9 k\end{array}\right]$
Thus, $\mathrm{x}=\frac{-9 k}{13}, \mathrm{y}=\frac{-k}{13}$ and $z=k$ (where $k$ is any real number ) satisfy the given system of equations