If the system of linear equations  $x+ky+3z=0,3x+ky-2z=02x+4y-3z=0$  has a non-zero solution  $\left(x,y,z\right)$  , then  $\frac{xz}{y^2}$ is equal to 

We have ,

$x+ky+3z=0;3x+ky-2z=0;2x+4y-3z=0$

System of equation has non-zero solution, if 

$\begin{array}{l}\left|\begin{array}{lcc}1& k& 3\\ 3& k& -2\\ 2& 4& -3\end{array}\right|=0\\ \Rightarrow (-3k+8)-k(-9+4)+3(12-2k)=0\\ \Rightarrow -3k+8+9k-4k+36-6k=0\\ \Rightarrow -4k+44=0\Rightarrow k=11\end{array}$

$\text{ Let }z=\lambda \text{, then we get }$

$\begin{array}{r}x+11y+3\lambda =0----\left(1\right)\\ 3x+11y-2\lambda =0-----\left(2\right)\end{array}$

$\text{ And }2x+4y-3\lambda =0----\left(3\right)$

Solving Eqs. (i) and (ii), we get 

$x=\frac{5\lambda }{2},y=\frac{-\lambda }{2},z=\lambda \Rightarrow \frac{xz}{y^2}=\frac{5{\lambda }^2}{2\times {\left(-\frac{\lambda }{2}\right)}^2}=10$