We have $\frac{AH}{2R}=|\cos A|=\cos A$ and similar equalities, because triangle is acute. Using the trigonometric identity in triangle: $\frac{229}{4R^2}-1=\cos^2A+\cos^2B+\cos^2C-1=-2\cos A\cos B\cos C=-2\cdot \frac{2\cdot12\cdot9}{8R^3}\implies 0=4R^3-229R-216=(R-8)(4R^2+32R+27)$ $32/4>0\wedge 27/4>0\implies (4x^2+32x+27=0\implies x<0)$ so $R=8$ $\sin A=\sqrt{1-\cos^2A}=3\sqrt7/8\wedge \sin B=\sqrt7/16\wedge \sin C=\sqrt3/2$ Area is equal to $2R^2\cdot \sin A\sin B \sin C=189\sqrt3/16$