Question:

Step 1: Establish key relations Let $R$ be the circumradius of $\triangle ABC$. Given $R=3$. For the orthocenter $H$, the distances from the orthocenter to the vertices are given by: $$AH = 2R|\cos A| \implies AH^2 = 4R^2 \cos^2 A$$ $$BH = 2R|\cos B| \implies BH^2 = 4R^2 \cos^2 B$$ $$CH = 2R|\cos C| \implies CH^2 = 4R^2 \cos^2 C$$ The side lengths of the triangle are related to the circumradius by the sine rule: $$BC = a = 2R \sin A \implies BC^2 = 4R^2 \sin^2 A$$ $$AC = b = 2R \sin B \implies AC^2 = 4R^2 \sin^2 B$$ $$AB = c = 2R \sin C \implies AB^2 = 4R^2 \sin^2 C$$ Step 2: Simplify the terms Substitute the relations from Step 1 into the terms of the expression: $$AH^2 + BC^2 = 4R^2 \cos^2 A + 4R^2 \sin^2 A = 4R^2 (\cos^2 A + \sin^2 A) = 4R^2$$ Similarly, for the other two terms: $$BH^2 + AC^2 = 4R^2 \cos^2 B + 4R^2 \sin^2 B = 4R^2 (\cos^2 B + \sin^2 B) = 4R^2$$ $$CH^2 + AB^2 = 4R^2 \cos^2 C + 4R^2 \sin^2 C = 4R^2 (\cos^2 C + \sin^2 C) = 4R^2$$ Step 3: Calculate the product Multiply the simplified terms: $$(AH^2 + BC^2)(BH^2 + AC^2)(CH^2 + AB^2) = (4R^2)(4R^2)(4R^2) = (4R^2)^3 = 64R^6$$ Step 4: Evaluate the given expression Substitute the product into the full expression: $$\frac{1}{64}(AH^2 + BC^2)(BH^2 + AC^2)(CH^2 + AB^2) = \frac{1}{64} \cdot 64R^6 = R^6$$ Step 5: Substitute R = 3 Given that the circumradius $R = 3$ units: $$R^6 = 3^6 = 729$$