jmc

The diagram the problem gives is drawn very out of scale so we redraw the diagram, this time with $\overline{AC}$ as the base:

All angles are given in degrees.

Let $\angle ECB=y$, so $\angle DBC=3y$. From $△AEC$ we have $\angle ACE=180^{\circ }-60^{\circ }-90^{\circ }=30^{\circ }$.

Now let $EC$ and $BD$ intersect at $F$. $\angle BFE=\angle DFC$ by vertical angles and $\angle BEF=\angle CDF=90^{\circ }$, so $\angle FBE=\angle FCD$, which is equal to 30 degrees. Now summing the angles in $△ABC$, we have $60^{\circ }+30^{\circ }+3y+y+30^{\circ }=180$, solving yields $4y=60$ so $y=15$ and we see $△BDC$ is a 45-45-90 triangle. Also, $△ABD$ is a 30-60-90 triangle.

Let $AD=x$, so $AB=2x$ and $DB=DC=x\sqrt{3}$. $BC=x\sqrt{3}\sqrt{2}=x\sqrt{6}$. We are given that this equals 12, so we find $x=12/\sqrt{6}=2\sqrt{6}$. It follows that the area of $△ABC$ can be found via $$(1/2)(AC)(BD)=(1/2)(x+x\sqrt{3})(x\sqrt{3})=12\sqrt{3}+36.$$ To find $EC$, notice that the area of $△ABC$ can also be written as $(1/2)(AB)(EC)$. Thus, $$(1/2)(4\sqrt{6})(EC)=12\sqrt{3}+36\Rightarrow EC=3(\sqrt{2}+\sqrt{6}).$$ Hence $a=3$, $b=2$, and $c=6$, so $a+b+c=\boxed{11}$.