smc

It is clear that $△ACE$ is an equilateral triangle. From the Law of Cosines on $△ABC$, we get that $A{C}^2=r^2+1^2-2r\cos \frac{2\pi }{3}=r^2+r+1$. Therefore, the area of $△ACE$ is $\frac{\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle. If we extend $BC$, $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$, $BC$ and $DE$ meet at $Y$, and $DE$ and $FA$ meet at $Z$, we find that hexagon $ABCDEF$ is formed by taking equilateral triangle $XYZ$ of side length $r+2$ and removing three equilateral triangles, $ABX$, $CDY$ and $EFZ$, of side length $1$. The area of $ABCDEF$ is therefore $\frac{\sqrt{3}}{4}(r+2{)}^2-\frac{3\sqrt{3}}{4}=\frac{\sqrt{3}}{4}(r^2+4r+1)$. Based on the initial conditions, $$\frac{\sqrt{3}}{4}(r^2+r+1)=\frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)$$ Simplifying this gives us $r^2-6r+1=0$. By Vieta's Formulas we know that the sum of the possible value of $r$ is $\boxed{6}$.