System of equations has non-zero solution when determinant of coefficient = 0.

So, in this questions,

$$\left| {\matrix{ 1 & K & 3 \cr 3 & K & { - 2} \cr 2 & 4 & { - 3} \cr } } \right| = 0$$

$$ \Rightarrow \,\,\,\,$$ ($$-$$ 3K + 8) $$-$$ K ($$-$$9 + 4) + 3(12 $$-$$ 2K) = 0

$$ \Rightarrow \,\,\,\,$$ $$-$$ 3K + 8 + 9K $$-$$ 4K + 36 $$-$$ 6K = 0

$$ \Rightarrow \,\,\,\,$$ $$-$$ 4K + 44 = 0

$$ \Rightarrow \,\,\,\,$$ K = 11

Now the equations become

x + 11y + 3z = 0 . . . (1)

3x + 11y $$-$$ 2z = 0 . . . (2)

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

By adding equation (1) and (3) we get,

3x + 15y = 0

$$ \Rightarrow \,\,\,\,$$ x = $$-$$ 5y

Putting x = $$-$$ 5y in equation (1) we get

$$-$$ 5y + 11y + 3z = 0

$$ \Rightarrow \,\,\,\,$$ 6y + 3z = 0

$$ \Rightarrow \,\,\,\,$$ z = $$-$$ 2y

$$\therefore\,\,\,\,$$ $${{xz} \over {{y^2}}}$$

$$ = {{\left( { - 5y} \right)\left( { - 2y} \right)} \over {{y^2}}}$$

$$ = {{10{y^2}} \over {{y^2}}}$$

$$ = 10$$