Given $\triangle ABC\ :\ A(0,a),B(-b,0),C(c,0)$ with area $\frac{1}{2}a(b+c)=1$. Points $A'(b+2c,0),B'(-c,2a),C'(-2b,-a)$. Distance $A'C'=\sqrt{(3b+2c)^{2}+a^{2}}$. Equation of $A'C'\:\ y=\frac{a}{3b+2c}[x-(b+2c)]$ or $ax-(3b+2c)y-a(b+2c)=0$, distance of the point $B'$ to this line $\left|\frac{-ac-2a(3b+2c)-a(b+2c)}{\sqrt{(3b+2c)^{2}+a^{2}}}\right|$. Area $\triangle A'B'C'=\frac{1}{2}a(7b+7c)=7$.

Alternatively, you can first show that the given green triangle has the same area with every single red ones, then show that every single red ones have the same area as every single blue ones, and the result follows.

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