Japan Mathematical Olympiad

\boxed{\frac{4\sqrt{21}}{7}}

Let $P(Q)$ be the point of intersection of the line $DE$ and the circum-circle of the triangle $ABC$, which lies on the opposite side (on the same side, respectively) as the point $O$ with respect to the line $AB$. Let $x=OD=OE$, $y=AE$ and $z=EC$. Then, we have $OP=OQ=OA=7$. From the similarity of the triangles $△ADQ$ and $△PDB$ it follows that $$AD\cdot BD=PD\cdot DQ=(OP-OD)(OQ+OD),$$ from which it follows that $8\cdot 3=(7-OD)(7+OD)=(7-x)(7+x)=49-x^2$ and therefore, we get $x^2=49-24=25$ and we have $x=5$. From the similarity of the triangles $△AEQ$ and $△PEC$, we also get $PE\cdot QE=AE\cdot CE$, from which we get $(7+5)(7-5)=yz$. Let now $\alpha =\angle AOD$, then $\angle AOE=180^{\circ }-\alpha$. Applying the law of cosine to the triangles $△AOD$ and $△AOE$, we get $$A{O}^2+D{O}^2-2AO\cdot DO\cos \alpha =A{O}^2,A{O}^2+E{O}^2-2AO\cdot EO\cos (180^{\circ }-\alpha )=y^2,$$ from which we get $$7^2+5^2-2\cdot 7\cdot 5\cos \alpha =8^2,7^2+5^2+2\cdot 7\cdot 5\cos \alpha =y^2,$$ and therefore, we have $$2(49+25)=64+y^2.$$ From this it follows that $y=\sqrt{84}=2\sqrt{21}$ and $EC=z=\frac{12\cdot 2}{y}=\frac{24}{2\sqrt{21}}=\frac{4}{7}\sqrt{21}$, which gives the desired answer to the problem.