12. (10 points) As shown in the figure, in triangle $A B C$, $A F=2 B F, C E=3 A E, C D=2 B D$, connect $C F$ intersecting $D E$ at point $P$, find the value of $\frac{\mathrm{EP}}{\mathrm{DP}}$.
【Analysis】As shown in the figure,
Connect $D F$. According to the given information, we can deduce that $\triangle B F D \cong \triangle B A C$, which leads to $\angle B D F=\angle B C A$. This implies $D F \parallel C A$, and $\frac{\mathrm{DF}}{\mathrm{AC}}=\frac{1}{3}$. We also find that $\frac{\mathrm{DF}}{\mathrm{CE}}=\frac{4}{9}$, and $\frac{\mathrm{CE}}{\mathrm{DF}}=\frac{9}{4}$. The result follows from the theorem of parallel lines dividing segments proportionally.
【Solution】Solution:

Connect $D F$,
Since $A F=2 B F$,
$$
C D=2 B D \text {, }
$$

Therefore, $\frac{B F}{A B}=\frac{B D}{B C}=\frac{1}{3}$,
Since $\angle B=\angle B$,
Therefore, $\triangle B F D \cong \triangle B A C$
Therefore, $\angle B D F=\angle B C A$
Therefore, $D F \parallel C A$,
$$
\frac{\mathrm{DF}}{\mathrm{AC}}=\frac{1}{3}
$$

Since $C E=3 A E$,
Therefore, $\frac{\mathrm{DF}}{\mathrm{CE}}=\frac{4}{9}$,
Therefore, $\frac{\mathrm{CE}}{\mathrm{DF}}=\frac{9}{4}$,
Since $D F \parallel C A$,
Therefore, $\frac{\mathrm{EP}}{\mathrm{DP}}=\frac{\mathrm{CE}}{\mathrm{DF}}=\frac{9}{4}$.