$17 \cdot 145$ As shown in the figure, in $\triangle ABC$, it is known that $BD: DC=3: 2$, $AE: EC=3: 4$, $M$ is the intersection of $AD$ and $BE$, and the area of $\triangle ABC$ is 1. Then, the area of $\triangle BMD$ is
(A) $\frac{3}{5}$.
(B) $\frac{3}{7}$.
(C) $\frac{12}{35}$.
(D) $\frac{4}{15}$.
(E) $\frac{4}{9}$.
[Solution] As shown in the figure, draw $E N / / A D$ intersecting $B C$ at $N$.
Since $\frac{A E}{E C}=\frac{D N}{N C}=\frac{3}{4}$, and $\frac{B D}{D C}=\frac{3}{2}$, then $N C: D N: B D=8: 6: 21$.
Since $S_{\triangle B E C}=\frac{4}{7} S_{\triangle A B C}=\frac{4}{7}$, we have
$$
S_{\triangle B N E}=\frac{27}{35} S_{\triangle B E C}=\frac{27 \cdot 4}{35 \cdot 7},
$$

Since $D M / / E N$, then $\triangle B D M \backsim \triangle B N E$. Therefore, $S_{\triangle B D M}=\left(\frac{21}{27}\right)^{2} S_{\triangle B N E}=\frac{49}{81} \cdot \frac{27 \cdot 4}{35 \cdot 7}=\frac{4}{15}$. Hence, the answer is $(D)$.