In $\triangle ABC$, point $E$ lies on $\overline{BC}$, point $F$ lies on $\overline{AC}$, and segments $\overline{AE}$ and $\overline{BF}$ intersect at $D$. The areas of $\triangle ADF$, $\triangle ADB$, and $\triangle BDE$ are $3$, $5$, and $6$, respectively. What is the area of quadrilateral $DECF$?

1. Connect $CD$, splitting $DECF$ into $[CDF]=u$ and $[CDE]=v$.
2. Triangles with the same height have areas proportional to their bases. Along $AE$: $\frac{AD}{DE}=\frac{[ABD]}{[BDE]}=\frac56$ and also $\frac{AD}{DE}=\frac{[ACD]}{[CDE]}=\frac{3+u}{v}$, so $6(3+u)=5v$.
3. Along $BF$: $\frac{BD}{DF}=\frac{[ABD]}{[ADF]}=\frac53$ and also $\frac{BD}{DF}=\frac{[BCD]}{[CDF]}=\frac{6+v}{u}$, so $5u=3(6+v)$.
4. Solving, $u=\frac{144}7$ and $v=\frac{198}7$, so $[DECF]=u+v=\boxed{\frac{342}7}$.