Three triangles and three circles

In triangle $LOA$ ,

$\cos 30=\dfrac{2}{LA}$ $\implies$ $\dfrac{\sqrt{3}}{2}=\dfrac{2}{LA}$ $\implies$ $LA=\dfrac{4}{\sqrt{3}}$

In triangle $AMP$ ,

$\cos 30=\dfrac{3}{MA}$ $\implies$ $\dfrac{\sqrt{3}}{2}=\dfrac{3}{MA}$ $\implies$ $MA=\dfrac{6}{\sqrt{3}}$

Apply cosine law in triangle $ALM$ ,

$(LM)^2=(LA)^2+(MA)^2-2(LA)(MA)(\cos 60)=\left(\dfrac{4}{\sqrt{3}}\right)^2+\left(\dfrac{6}{\sqrt{3}}\right)^2-2\left(\dfrac{4}{\sqrt{3}}\right)\left(\dfrac{6}{\sqrt{3}}\right)\left(\dfrac{1}{2}\right)=\dfrac{16}{3}+12-8=\dfrac{28}{3}$

Since triangle $MNL$ is equilateral (the proof is left to the solver), the area is

$A=\dfrac{1}{2}(LM)^2(\sin 60)=\dfrac{1}{2}\left(\dfrac{28}{3}\right)\left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{7}{3}\sqrt{3}$

The desired answer is $ab=7(3)=$ $\boxed{21}$