Too Many Circles, Too Distracting

Let $I_c$ be the $C$ -excenter of triangle $ABC$ . Since the internal and angle bisectors of any angle are perpendicular, it follows that triangle $ABC$ is the orthic triangle of triangle $I_aI_bI_c$ . Therefore, the circumradius of triangle $II_aI_b$ is the circumradius of triangle $I_aI_bI_c$ , which is twice the circumradius of triangle $ABC$ . By the Law of Cosines, $\cos A=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{8^2+5^2-7^2}{2\cdot 8\cdot 5}=\dfrac{1}{2}$ Therefore, $\sin A=\dfrac{\sqrt{3}}{2}$ . By the Extended Law of Sines, the circumradius of triangle $ABC$ is $\dfrac{a}{2\sin A}=\dfrac{7}{\sqrt{3}}$ . It follows that $\boxed{\sqrt{3}R=14}$ .