In triangle ABC, E is a point on AC and F is a point on AB. BE and CF intersect at D. If the areas of triangles BDF, BCD and CDE are 3, 7 and 7 respectively, what is the area of the quadrilateral AEDF?

I tried to solve the question as the ratios of the areas of pentagon EDFBC and the area of the quadrilateral AEDF. Area of pentagon = 17. I couldn't proceed from here. Please advise.

Use the notation $[UVW]$ to denote the area of triangle $UVW$.
Draw line segment $AD$.
Let $x=[ADF]$, and let $y=[ADE]$.
Applying the key principle, we get $$\frac{x+3}{y}=[ADB]:[ADE]=BD:DE= [BCD]:[CDE] = 7:7 = 1:1$$ and also $$\frac{x}{y+7}=[ADF]:[ADC]=FD:DC= [BDF]:[BCD] = 3:7 \qquad\;\;\;\;\;$$ Solving the system $$\begin{cases} {\Large{\frac{x+3}{y}}}=1\\[4pt] {\Large{\frac{x}{y+7}}}={\large{\frac{3}{7}}}\\ \end{cases}$$ for $x,y$ yields $$x=\frac{15}{2},\;\;\;y=\frac{21}{2}$$ so the area of quadrilateral $AEDF$ is $x+y=18$.