B. As shown in Figure 3, given that $D$ and $E$ are points on the sides $AC$ and $AB$ of $\triangle ABC$, respectively, and line $BD$ intersects $CE$ at point $F$. If the areas of $\triangle CDF$, $\triangle BFE$, and $\triangle BCF$ are $3$, $4$, and $5$, respectively, then the area of quadrilateral $AEFD$ is $\qquad$
B. $\frac{204}{13}$.

Connect $A F$. Then
$$
\begin{array}{l}
\frac{S_{\triangle A E F}+4}{S_{\triangle A F D}}=\frac{S_{\triangle A E F}+S_{\triangle B F E}}{S_{\triangle A F D}} \\
=\frac{B F}{F D}=\frac{S_{\triangle A C F}}{S_{\triangle C D F}}=\frac{5}{3}, \\
\frac{S_{\triangle A F D}+3}{S_{\triangle A E F}}=\frac{S_{\triangle A F D}+S_{\triangle C D F}}{S_{\triangle A E F}} \\
=\frac{C F}{F E}=\frac{S_{\triangle A C F}}{S_{\triangle A E F}}=\frac{5}{4} .
\end{array}
$$

Solving, we get $S_{\triangle A E F}=\frac{108}{13}, S_{\triangle A F D}=\frac{96}{13}$.
Therefore, the area of quadrilateral $A E F D$ is $\frac{204}{13}$.