Solution:

Since $\angle OIA = 90^\circ,$ the Pythagorean theorem in triangle $OIA$ gives $IA^2 = OA^2 - OI^2 = R^2 - OI^2,$ and Euler's formula $OI^2 = R^2 - 2Rr$ yields $$IA^2 = 2Rr = 2 \cdot 13 \cdot 6 = 156.$$ Combining with $IA = \frac{r}{\sin(A/2)}$ gives $\sin^2\frac{A}{2} = \frac{36}{156} = \frac{3}{13},$ so $\cos^2\frac{A}{2} = \frac{10}{13}.$

Then $\sin A = 2 \sin\frac{A}{2}\cos\frac{A}{2} = \frac{2\sqrt{30}}{13},$ so $a = BC = 2R \sin A = 4\sqrt{30},$ while $s - a = r \cot\frac{A}{2} = 6\sqrt{\frac{10}{3}} = 2\sqrt{30}.$ Hence the semiperimeter is $s = 6\sqrt{30}.$

Equating the two area formulas $[ABC] = rs = \frac{1}{2}\, bc \sin A,$ $$bc = \frac{2rs}{\sin A} = \frac{2 \cdot 6 \cdot 6\sqrt{30}}{2\sqrt{30}/13} = 36 \cdot 13 = 468.$$