4. In $\triangle A B C$, $B C=2, C A=3, A B=4, P$ is a point inside $\triangle A B C$, $D, E, F$ are on $A B, B C, C A$ respectively, and $P D \parallel B C, P E \parallel A C, P F \parallel A B$. If $P D=P E=P F=l$, then $l=$
4. Connect $A P, B P, C P$ and extend them to intersect $B C, C A, A B$ at $G, H, K$ respectively. Since $P D / / B C$, we have $\frac{P D}{B C}=\frac{P K}{C K}=\frac{S_{\triangle B P K}}{S_{\triangle B C K}}=\frac{S_{\triangle A P K}}{S_{\triangle A C K}}$. By the property of equal ratios, we have $\frac{l}{2}=\frac{S_{\triangle B P K}+S_{\triangle A P K}}{S_{\triangle B C K}+S_{\triangle A C K}}=\frac{S_{\triangle A B P}}{S_{\triangle A B C}}$.
Similarly, $\frac{l}{3}=\frac{S_{\triangle B C P}}{S_{\triangle A B C}}$,
$$
\frac{l}{4}=\frac{S_{\triangle A C P}}{S_{\triangle A B C}}
$$

Thus, $\frac{l}{2}+\frac{l}{3}+\frac{l}{4}$
$$
\begin{array}{l}
=\frac{S_{\triangle A B P}+S_{\triangle B C P}+S_{\triangle A C P}}{S_{\triangle A B C}} \\
=1 .
\end{array}
$$

Therefore,
$$
l=\frac{1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}=\frac{12}{13} .
$$