The equations $x-2 y-3=0$ and $18 x-k^{2} y-9 k=0$ represent two lines. For some real number $k$, these two lines are distinct and parallel. Determine the value of $k$.
We can re-write the equation of the first line in slope-intercept form as $y=\frac{1}{2} x-\frac{3}{2}$.

If $k=0$, then the second line has equation $18 x=0$ or $x=0$ which is vertical and cannot be parallel to the first line which has slope $\frac{1}{2}$.

This means $k \neq 0$, so the equation of the second line can be rearranged to $y=\frac{18}{k^{2}} x-\frac{9}{k}$.

In order for these lines to be parallel, they must have the same slope.

Therefore, $\frac{1}{2}=\frac{18}{k^{2}}$, so $k^{2}=36$ or $k= \pm 6$.

If $k=6$, then the equation of the second line is $y=\frac{18}{36} x-\frac{9}{6}$ or $y=\frac{1}{2} x-\frac{3}{2}$, which is the same equation as that of the other line.

Since the lines are distinct, we conclude that $k=-6$.

Indeed, the lines with equations $y=\frac{1}{2} x-\frac{3}{2}$ and $y=\frac{1}{2} x+\frac{3}{2}$ are parallel and distinct.

ANSWER: -6