After rationalizing the numerator of $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}$, the denominator in simplest form is:
$\textbf{(A)}\ \sqrt{3}(\sqrt{3}+\sqrt{2})\qquad\textbf{(B)}\ \sqrt{3}(\sqrt{3}-\sqrt{2})\qquad\textbf{(C)}\ 3-\sqrt{3}\sqrt{2}\qquad\\ \textbf{(D)}\ 3+\sqrt6\qquad\textbf{(E)}\ \text{None of these answers}$
To rationalize the numerator, multiply by the conjugate.
\[\frac{\sqrt3 - \sqrt2}{\sqrt3}\cdot \frac{\sqrt3 + \sqrt2}{\sqrt3 + \sqrt2} = \frac{1}{3+\sqrt6}.\]
The denominator is $\boxed{\mathrm{(D)}\text{ } 3+\sqrt6 }.$