jmc

We will also adopt the notation $|△XYZ|$ to represent the area of $△XYZ$.

Recall that if two triangles have their bases along the same straight line and they share a common vertex that is not on this line, then the ratio of their areas is equal to the ratio of the lengths of their bases.

Using this fact, $$\frac{|△AEF|}{|△DEF|}=\frac{AF}{FD}=\frac{3}{1}.$$Thus, $$|△AEF|=3\times |△DEF|=3(17)=51.$$Then, $$|△AED|=|△AEF|+|△DEF|=51+17=68.$$Also, $$\frac{|△ECD|}{|△AED|}=\frac{EC}{AE}=\frac{2}{1}.$$Thus, $$|△ECD|=2\times |△AED|=2(68)=136.$$Then, $$|△DCA|=|△ECD|+|△AED|=136+68=204.$$Since $D$ is the midpoint of $BC$, $$\frac{|△BDA|}{|△DCA|}=\frac{BD}{DC}=\frac{1}{1}.$$Then, $|△BDA|=|△DCA|=204$ and $$|△ABC|=|△BDA|+|△DCA|=204+204=\boxed{408}.$$