Geometry problem about two externally touching circles

Let $PQ=QR=RS=2x$, $AM$ and $O_2N$ be perpendiculars to $PS$

and $O_2K$ be a perpendicular to $AM$.

Thus, since $$KO_2=MN=x+2x+x=4x,$$ $$AK=\sqrt{12^2-x^2}-\sqrt{3^2-x^2}$$ and $$AO_2=12+3=15,$$ by the Pythagoras's theorem for $\Delta AO_2K$ we obtain: $$(4x)^2+\left(\sqrt{12^2-x^2}-\sqrt{3^2-x^2}\right)^2=15^2.$$ Can you end it now?