12. (10 points) As shown in the figure, in triangle $A B C$, $D$ is the midpoint of $B C$, $A F=2 B F$, $C E=3 A E$, connect $C F$ intersecting $D E$ at point $P$, find the value of $\frac{\mathrm{EP}}{\mathrm{DP}}$.
【Analysis】Connect $E F, D F$. It is easy to know that $E P: D P=S_{\triangle E F C}: S_{\triangle D F C}$, solving for $S_{\triangle E F C}: S_{\triangle D F C}$ will solve the problem.

【Solution】Connect $E F, D F$.
$$
\begin{array}{l}
\because E P: D P=S_{\triangle E F C}: S_{\triangle D F C}, \\
\text { Also } \because S_{\triangle D F C}=\frac{1}{2} S_{\triangle B F C}, S_{\triangle E F C}=\frac{3}{4} S_{\triangle A F C}, S_{\triangle F B C}=\frac{1}{2} S_{\triangle A F C}, \\
\therefore E P: \quad D P=\frac{3}{4} S_{\triangle A F C}: \frac{1}{2} S_{\triangle B F C}=\frac{3}{4} S_{\triangle A F C}: \frac{1}{4} S_{\triangle A F C}=3: 1, \\
\therefore \frac{E P}{\mathrm{DP}}=3 .
\end{array}
$$