$17 \cdot 92$ In a scalene triangle, the medians $A N$ and $B P$ are 3 and 6 units long, respectively, and the area is $3 \sqrt{15}$. The length of the third median $C M$ is
(A) 4.
(B) $3 \sqrt{3}$.
(C) $3 \sqrt{6}$.
(D) $6 \sqrt{3}$.
(E) $6 \sqrt{6}$.
[Solution] Extend $G M$ to $D$ such that $M D=G M$, and connect $A D, B D$. Then quadrilateral $A D B G$ is a parallelogram.
Then $\frac{1}{2} S_{\triangle A D B G}=S_{\triangle A B G}=\frac{1}{3} S_{\triangle A B C}=\sqrt{15}$,
and $\frac{1}{2} S_{\square A D B G}=S_{\triangle A G D}$
$$
=\sqrt{p(p-u)(p-v)(p-t)},
$$

where $u, v, t$ are the three side lengths of $\triangle A G D$, and $p$ is its semi-perimeter.
Assume $u=2 x, v=4, t=2, p=3+x$, substituting into the above formula gives
$$
\sqrt{(3+x)(3-x)(x-1)(x+1)}=\sqrt{15} \text {, }
$$

which simplifies to
$$
\left(x^{2}-1\right)\left(9-x^{2}\right)=15 \text {, }
$$

or
$$
x^{4}-10 x^{2}+24=0 \text {. }
$$

Solving this, we get $x^{2}=4$ or 6, i.e., $x=2$ or $\sqrt{6}$.
Thus, $C M=6$ or $3 \sqrt{6}$, but since $B P=6$, and the triangle is not isosceles, $C M=6$ is discarded.
Therefore,
$$
C M=3 \sqrt{6} .
$$

Hence, the answer is $(C)$.