For a homogeneous system to have a non-zero solution, the determinant of its coefficient matrix must be zero.

The system is: $x+ky+3z=0$ $3x+ky-2z=0$ $2x+4y-3z=0$

So the coefficient matrix is $A=\begin{pmatrix}1&k&3\\3&k&-2\\2&4&-3\end{pmatrix}$

Set $\det(A)=0$.

$\det\begin{pmatrix}1&k&3\\3&k&-2\\2&4&-3\end{pmatrix}=0$

Expanding along the first row: $=1\begin{vmatrix}k&-2\\4&-3\end{vmatrix}-k\begin{vmatrix}3&-2\\2&-3\end{vmatrix}+3\begin{vmatrix}3&k\\2&4\end{vmatrix}$

$=1(k\cdot(-3)-(-2)\cdot4)-k(3\cdot(-3)-(-2)\cdot2)+3(3\cdot4-2k)$

$=(-3k+8)-k(-9+4)+3(12-2k)$

$=(-3k+8)+5k+36-6k$

$=44-4k$

Therefore, $44-4k=0 \implies k=11$

Substitute $k=11$ into the system: $x+11y+3z=0 \quad ...(1)$ $3x+11y-2z=0 \quad ...(2)$ $2x+4y-3z=0 \quad ...(3)$

Eliminate variables to find the ratio $\dfrac{xz}{y^2}$.

Subtract (1) from (2): $2x-5z=0 \implies 2x=5z \implies x=\frac{5z}{2}$

Put this in (3): $2\left(\frac{5z}{2}\right)+4y-3z=0$ $5z+4y-3z=0$ $2z+4y=0$ $z=-2y$

Then $x=\frac{5z}{2}=\frac{5(-2y)}{2}=-5y$

Now compute: $\frac{xz}{y^2}=\frac{(-5y)(-2y)}{y^2}=\frac{10y^2}{y^2}=10$

Hence the correct option is: $\boxed{10}$ which is Option C.