(Color+beauti)ful Hexagon!

If each side is $a$ , then $BP= \dfrac{a}{2}$ , $PQ=\dfrac{a}{2\sqrt{3}}$ , $AC=a \sqrt{ 3}$ and $A(\triangle BCD) = A(\triangle ABC)$ and now its easy.

First note that $\Delta BCD$ is congruent to $\Delta ABC.$ Now $\angle BCD = 120^{\circ}$ , so as $\Delta BCD$ is isosceles we have that $\angle CBD = 30^{\circ},$ so $\angle PBQ = \angle PBC - \angle CBD = 30^{\circ}.$

The ratio of the areas of $\Delta PBQ$ to $\Delta PBC$ is then $\frac{\tan(30^{\circ})}{\tan(60^{\circ})} = \dfrac{1}{3}.$

Then since the area of $\Delta PBC$ is $\frac{1}{2}$ that of $\Delta ABC$ , and hence of $\Delta BCD$ as well, the ratio of the areas of the green triangle to the red triangle is $1 : 6.$

Next, let $O$ be the center of the hexagon. Then $\angle BOD = \angle BCD = 120^{\circ},$ and thus the red triangle is $\frac{1}{2}$ the area of quadrilateral $OBCD.$ But the area of $OBCD$ is $\frac{1}{3}$ that of the hexagon, and thus the area of the red triangle is $\frac{1}{6}$ that of the hexagon.

So now suppose the area of the hexagon is $36.$ The area of the red triangle would then be $6$ , and hence the area of the green triangle $1.$ This would then give us an area for the purple region of $36 - 6 - 1 = 29.$ Thus the desired ratio of areas of colors purple:green:red is $29:1:6$ , and so $a + b + c = 29 + 1 + 6 = \boxed{36}.$

$\Delta~BCD~\equiv~\Delta~ABC ....(1)~~~\therefore ~ \angle CBD (i.e.~\angle ~CBQ )= \angle~BAC...(2)\\\text{Let AB=BC=CD=DE=1. }~~~~~~~~In~\Delta ABC,~AB=BC, \angle ABC=120^o\\\therefore \angle BAC=\angle ACB=30^o =\angle QBC~~~by~~~(2)\\\because of symmetry, \Delta ~PBA~\equiv~\Delta~PBC . ~\therefore~\Delta PBC ~is~a~90:60:30....\color{#D61F06}{(3)}\\\angle PBC=60^o.....(4)~~~~\angle PBQ=\angle PBC-\angle QBC=30^o.\\In ~ \Delta~ QBP,~\angle QPB=90^o,~\angle PBQ=30^o~\therefore ~This~is~60:30:90~ \Delta .\color{#3D99F6}{(5)}\\In~a~30^o:60^o:90^o~~\Delta, ~~AREA=\dfrac{Hyp.^2 *\sqrt3}{8}=\dfrac{(Big~leg)^2 *\sqrt3}{6}.\\ Side=1, hexagonal ~area=\dfrac{3*\sqrt3}{2}......\color{#69047E}{(6)} \\From~~(5)~~\color{#3D99F6}{area~\Delta PBQ =\dfrac{BP^2 *\sqrt3}{6}= \dfrac{(\dfrac{1}{2})^2 *\sqrt3}{6 }=\large \dfrac{\sqrt3}{24} }\\From~~(3)~~area~\Delta PBC =\dfrac{BC^2 *\sqrt3}{8}= \dfrac{\sqrt3}{8} =\dfrac{area~\Delta ABC}{2}\\=\dfrac{area~\Delta BCD}{2}...from~(1).~~\therefore~\color{#D61F06}{Area~of~\Delta BCD=\large \dfrac{\sqrt3}{4} }\\Purple~area=hexagonal~area~-red~area~-~green~area.\\\large =\color{#69047E}{\dfrac{3*\sqrt3}{2}-\dfrac{\sqrt3}{4}- \dfrac{\sqrt3}{24}.}\\ \{\dfrac{3*\sqrt3}{2}-\dfrac{\sqrt3}{4}- \dfrac{\sqrt3}{24}\}:\dfrac{\sqrt3}{4}: \dfrac{\sqrt3}{24}::\{36-1-6\}:1:6\\\therefore \large \{36-1-6\}+1+6= \boxed{36} \\~~\\$
$\large \text{Thank you, Mr. Brian, the equivalence of ABC and BCD, }\\ \large \text{was taken from your solution. }\\\large \text{That avoided construction of Altitude in BCD.}$