Solution:

Since $\ell \parallel BC,$ triangle $AEF$ is equilateral; let $s = AF = EF.$ The distance between $\ell$ and $BC$ is the height of $ABC$ minus the height of $AEF,$ so $BD = \frac{\sqrt{3}}{2}(840 - s);$ write $h = BD.$

In triangle $BED,$ the base $\overline{BD}$ is perpendicular to $BC$ and has length $h,$ while $E$ lies at horizontal distance $\frac{h}{\sqrt{3}}$ from line $BD$ because $\angle EBC = 60^\circ.$ Hence $[BED] = \frac{h^2}{2\sqrt{3}}.$ Also $\angle AFG = 180^\circ - \angle AFE = 120^\circ,$ and an isosceles triangle with a $120^\circ$ angle must have it as the apex angle, so $FA = FG = s$ and $[AFG] = \frac{1}{2}s^2 \sin 120^\circ = \frac{\sqrt{3}}{4}s^2.$

The ratio condition gives $$\frac{[AFG]}{[BED]} = \frac{\sqrt{3}s^2/4}{h^2/(2\sqrt{3})} = \frac{3}{2} \cdot \frac{s^2}{h^2} = \frac{8}{9},$$ so $\frac{s}{h} = \frac{4}{3\sqrt{3}}.$ Substituting $h = \frac{\sqrt{3}}{2}(840 - s)$ yields $s = \frac{2}{3}(840 - s),$ so $5s = 1680$ and $AF = 336.$