It's all Hexagons

Using the law of sines on $\triangle{ABC}$ we obtain:

$4h^{*} = \dfrac{a}{\sin(\alpha)} \implies \sin(\alpha) = \dfrac{a}{4h^{*}} \implies h^{*} = \dfrac{ab}{4h^{*}} \implies b = \dfrac{4{h^{*}}^{2}}{a}$ .

Using the law of cosines $\implies a^2 + ab + b^2 = 12{h^{*}}^2 \implies a^2 + 4{h^{*}}^2 + \dfrac{16{h^{*}}^4}{a^2} = 12{h^{*}}^2 \implies a^4 - 8{h^{*}}^2 a^2 + 16 {h^{*}}^4 = 0 \implies$

$(a^2 - 4{h^{*}}^2)^2 = 0 \implies a = 2h^{*} \implies b = 2h^{*} \implies a + b = 4h^{*} \implies$

the area of the larger hexagon $A = 24\sqrt{3}{h^{*}}^2 = {h^{*}}^{4} + 432 \implies {h^{*}}^{4} - 24\sqrt{3}{h^{*}}^2 + (12\sqrt{3})^2 = 0 \implies ({h^{*}}^2 - 12\sqrt{3})^2 = 0 \implies {h^{*}}^2 = 12\sqrt{3}$ $\implies h = 2\sqrt[\scriptstyle 4]{27}$

$\implies$ The sum of the areas of the red triangles $\sum_{j = 1}^{6} (A_{red})_{j} = 6(\dfrac{1}{2})(2\sqrt{3}{h^{*}}^2) = 6\sqrt{3}{h^{*}}^2 = 6\sqrt{3}(12\sqrt{3}) = 6 * 12 * 3 = \boxed{216}$ .