$\left| {\begin{array}{*{20}{c}} 1&k&3\\ 3&k&{ - 2}\\ 2&4&{ - 3} \end{array}} \right| = 0$

$ \Rightarrow k = 11$ 

Now equations become 

$x+11y+3z=0$           ......$(1)$

$3x+11y-2z=0$           ......$(2)$

$2x+4y-3z=0$              .......$(3)$

Adding equations $(1)$ and $(3)$  we get

$3x + 15y = 0 \Rightarrow k = 5 - y$

Now put $x=-5y$ in equation $(1)$, we get 

$ - 5y + 11y + 3z = 0 \Rightarrow z =  - 2y$

$\therefore \frac{{xz}}{{{y^2}}} = \frac{{\left( { - 5y} \right)\left( { - 2y} \right)}}{{{y^2}}} = 10$