\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 $\triangle ADQ$ and $\triangle PDB$ 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 $\triangle AEQ$ and $\triangle PEC$, 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 $\triangle AOD$ and $\triangle AOE$, we get
$$AO^2 + DO^2 - 2AO \cdot DO \cos \alpha = AO^2, \quad AO^2 + EO^2 - 2AO \cdot EO \cos(180^\circ - \alpha) = y^2,$$
from which we get
$$7^2 + 5^2 - 2 \cdot 7 \cdot 5 \cos \alpha = 8^2, \quad 7^2 + 5^2 + 2 \cdot 7 \cdot 5 \cos \alpha = y^2,$$
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.