There's a "straightforward" vector solution:
Let $AB=b$, $AD=d$ the basis vectors and $[x\times y]$ be the cross product, since we're given $|[AB\times AD]|= 70$.
Vectors $AF=1/3AB=1/3b, AG=2/3AB=2/3b, AE=AD+1/2DE=AD+1/2AB=d+1/2b$.
$X$ lies on the line $YZ$ iff $AX=t\cdot AY + (1-t)\cdot AZ$, where $t$ is a real.
So, consider
$AH=(1-u)\cdot 0+u\cdot (b+d)=(1-v)\cdot AE+v\cdot AF$
$u\cdot (b+d)=(1-v)\cdot (d+1/2b)+v/3\cdot b$
$u=(1-v)/2+v/3, u=(1-v) \Rightarrow u=2/5, v=3/5 \Rightarrow AH=2/5\cdot(b+d)$
$S_{\Delta AHF} = \frac{1}{2}[AH\times AF]= \frac{1}{2}[2/5\cdot(b+d)\times 1/3b]= \frac{1}{15}[(b+d)\times b] = \frac{1}{15}\left([b\times b] + [d\times b] \right)= \frac{1}{15} (0+70) = \frac{14}{3}$.
More classical way:
Triangle $AHF$ is similar to triangle $CHE$ with ratio $2:3$
That is all one needs. $AH/HC=2/3$, so $AH/AC=2/5$ and the "height" of $\Delta AHF$ is $2/5AD$, while its base is $1/3AB$ , $S=\frac{1}{2}\cdot\frac{2}{5}\cdot\frac{1}{3}\cdot 70=\frac{14}{3}$