For non-zero solution, $\Delta=0$:

$\begin{vmatrix}1&k&3\\3&k&-2\\2&4&-3\end{vmatrix}=1(−3k+8)−k(−9+4)+3(12−2k)=−3k+8+5k+36−6k=44−4k=0$

So $k=11$.

With $k=11$: from equations 1 and 2: $x+11y+3z=0$ and $3x+11y-2z=0$. Subtracting: $2x-5z=0\Rightarrow x=\frac{5z}{2}$.

Substituting back: $\frac{5z}{2}+11y+3z=0\Rightarrow 11y=-\frac{11z}{2}\Rightarrow y=-\frac{z}{2}$.

$\frac{xz}{y^2}=\frac{\frac{5z}{2}\cdot z}{\frac{z^2}{4}}=\frac{\frac{5z^2}{2}}{\frac{z^2}{4}}=\frac{5}{2}\times4=\mathbf{10}$