6. As shown in Figure 2, let $O$ be a point inside $\triangle ABC$. Connect $AO$, $BO$, and $CO$, and extend them to intersect $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Given that $S_{\triangle AOB}: S_{\triangle BOC}: S_{\triangle AOC}=$ 3:4:6. Then $\frac{OD}{AO} \cdot \frac{OE}{BO} \cdot \frac{OF}{CO}$ equals ( ).
(A) $\frac{2}{35}$
(B) $\frac{4}{35}$
(C) $\frac{6}{35}$
(D) $\frac{8}{35}$
6. B.

Since $\frac{S_{\triangle A O B}}{S_{\triangle A B C}}=\frac{3}{13}, \frac{S_{\triangle B O C}}{S_{\triangle A B C}}=\frac{4}{13}, \frac{S_{\triangle A O C}}{S_{\triangle A B C}}=\frac{6}{13}$, then $\frac{O F}{C F}=\frac{3}{13}, \frac{O D}{A D}=\frac{4}{13}, \frac{O E}{B E}=\frac{6}{13}$. Therefore, $\frac{O F}{C O}=\frac{3}{10}, \frac{O D}{A O}=\frac{4}{9}, \frac{O E}{B O}=\frac{6}{7}$.
Thus, $\frac{O D}{A O} \cdot \frac{O E}{B O} \cdot \frac{O F}{C O}=\frac{4}{35}$.