6. Let $A N$, $B P$, and $C M$ be the medians of $\triangle A B C$ with unequal sides, $A N=3$, $B P=6$, and $S_{\triangle A B C}$ $=3 \sqrt{15}$. Then $C M=(\quad)$.
(A)6
(B) $3 \sqrt{3}$
(C) $3 \sqrt{6}$
(D) $6 \sqrt{3}$
6. C.

As shown in Figure 4, let the centroid of $\triangle ABC$ be $G$, and $CM = 3x$. Then $CG = 2x$, $GM = x$.

Extend $CM$ to point $Q$ such that $MQ = GM$. Connect $BQ$ and $AQ$. Then quadrilateral $AQB G$ is a parallelogram.
Thus, $AQ = BG = 4$.
Also, $S_{\triangle AQC} = S_{\triangle ABG} = \frac{1}{3} S_{\triangle ABC} = \sqrt{15}$.
In $\triangle AQG$, $AG = 2$, $AQ = 4$, $QG = 2x$, and the semi-perimeter $p = \frac{1}{2}(2 + 4 + 2x) = 3 + x$.
By Heron's formula,
$\sqrt{15} = \sqrt{(3 + x)(3 - x)(x - 1)(x + 1)}$.
Simplifying, we get $x^4 - 10x^2 + 24 = 0$.
Solving, we find $x^2 = 4$ or $x^2 = 6$.
Thus, $x = 2$ or $x = \sqrt{6}$.
When $x = 2$, $CM = 6$.
Given $BP = 6$, it follows that $BA = AC$.
This contradicts the given conditions.
Therefore, $x = \sqrt{6}$, and $CM = 3\sqrt{6}$.