jmc

Using the identity $\cos A+\cos B+\cos C=1+\frac{r}{R}$, we have that $\cos A+\cos B+\cos C=\frac{21}{16}$. From here, combining this with $2\cos B=\cos A+\cos C$, we have that $\cos B=\frac{7}{16}$ and $\sin B=\frac{3\sqrt{23}}{16}$. Since $\sin B=\frac{b}{2R}$, we have that $b=6\sqrt{23}$. By the Law of Cosines, we have that:$$b^2=a^2+c^2-2ac\cdot \cos B⟹a^2+c^2-\frac{7ac}{8}=36\cdot 23.$$But one more thing: noting that $\cos A=\frac{b^2+c^2-a^2}{2cb}$. and $\cos C=\frac{a^2+b^2-c^2}{2ab}$, we know that $\frac{36\cdot 23+b^2+c^2-a^2}{bc}+\frac{36\cdot 23+a^2+b^2-c^2}{ab}=\frac{7}{4}⟹$ $\frac{36\cdot 23+c^2-a^2}{c}+\frac{36\cdot 23+a^2-c^2}{a}=\frac{21\sqrt{23}}{2}⟹$ $\frac{(a+c)(36\cdot 23+2ac-c^2-a^2)}{ac}=\frac{21\sqrt{23}}{2}$. Combining this with the fact that $a^2+c^2-\frac{7ac}{8}=36\cdot 23$, we have that: $\frac{(a+c)(-2ac\cdot \frac{7}{16}+2ac)}{ac}=\frac{21\sqrt{23}}{2}⟹$ $a+c=\frac{28\sqrt{23}}{3}$. Therefore, $s$, our semiperimeter is $\frac{23\sqrt{23}}{3}$. Our area, $r\cdot s$ is equal to $\frac{115\sqrt{23}}{3}$, giving us a final answer of $\boxed{141}$.