3. As shown in Figure 2, in the equilateral $\triangle ABC$, points $D, E, F$ on sides $BC, CA$, and $AB$ satisfy $BD=2CD, CE=3AE, AF=4BF$. If $S_{\triangle ABC}=36$, then the area of $\triangle LMN$ formed by lines $AD, BE, CF$ is . $\qquad$
3. $\frac{529}{52}$.

Consider any $\triangle XYZ$, with points $S$ and $T$ on sides $YZ$ and $XZ$ respectively, such that $YS = uSZ$ and $ZT = vTX$.

Let $YT$ and $XS$ intersect at point $K$, as shown in Figure 4.
By Menelaus' Theorem, we have
$$
\begin{array}{l}
\frac{YS}{SZ} \cdot \frac{ZX}{XT} \cdot \frac{TK}{KY} = 1 \\
\Rightarrow \frac{YK}{KT} = u(1 + v). \\
\frac{BM}{LE} = 2(1 + 3) = 8, \\
\frac{BM}{ME} = \frac{1}{4}\left(1 + \frac{1}{3}\right) = \frac{1}{3}, \\
\frac{CM}{MF} = 3(1 + 4) = 15, \\
\frac{CN}{NF} = \frac{1}{2}\left(1 + \frac{1}{4}\right) = \frac{5}{8}.
\end{array}
$$

Therefore, $BM : ML : LE = 9 : 23 : 4$,
$$
CN : NM : MF = 80 : 115 : 13.
$$

Thus, $S_{\triangle MN} = \frac{ML}{BE} \cdot \frac{MN}{MC} S_{\triangle BCE}$
$$
\begin{array}{l}
= \frac{ML}{BE} \cdot \frac{MN}{MC} \cdot \frac{CE}{CA} S_{\triangle ABC} \\
= \frac{23}{36} \times \frac{115}{195} \times \frac{3}{4} \times 36 = \frac{529}{52}.
\end{array}
$$