The system of linear equations $x+ky+3z=0$  ,  $3x+ky-2z=0$ and  $2x+4y-3z=0$  has a non-zero  solution (x,y,z) then $\frac{xz}{y^2}=$  
 

Since given system of linear equations $x+ky+3z=0$  , $3x+ky-2z=0$  and $2x+4y-3z=0$  has a non-zero solution then   
$\left|\begin{array}{ccc}1& K& 3\\ 3& K& -2\\ 2& 4& -3\end{array}\right|=0\Rightarrow (-3K+8)-K(-9+4)+3(12-2K)=0$
$\Rightarrow -3K+8+5K+36-6K=0$

$\Rightarrow +4K=33\Rightarrow K=11$
Let   $Z=\lambda$
 $x+11y+3\lambda =0$  ……(1)
 $2x+4y-3\lambda =0$………(2)
(1) + (2) 
$\Rightarrow 3x=-15y$
$\Rightarrow x=-5y$
From  (1) ,    $-5y+11y=-3x$
$\Rightarrow 6y=-3\lambda$
$\therefore y=\frac{-\lambda }{2}$
$\therefore x=\frac{5\lambda }{2}$
Hence,    $\frac{xz}{y^2}=\frac{\frac{5}{2}{\lambda }^2}{\frac{{\lambda }^2}{4}}=10$