Question:

**Step 1: Use properties of the triangle** Let the angles be $A > B > C$. We are given $A - C = \pi/2$. The circumradius is $R = 1$.\nThe sides are in A.P., so $2b = a + c$. By the Law of Sines, $a = 2R \sin A = 2 \sin A$, $b = 2 \sin B$, and $c = 2 \sin C$.\nSubstitute into the A.P. condition: $4 \sin B = 2 \sin A + 2 \sin C \implies 2 \sin B = \sin A + \sin C$. **Step 2: Solve for the angles** Using sum-to-product formula: $2 \sin B = 2 \sin\left(\frac{A+C}{2}\right) \cos\left(\frac{A-C}{2}\right)$.\nSince $A+C = \pi - B$, $\sin\left(\frac{A+C}{2}\right) = \cos(B/2)$.\nAlso, $\cos\left(\frac{A-C}{2}\right) = \cos(\pi/4) = \frac{1}{\sqrt{2}}$.\nThus, $2 \sin B = \sqrt{2} \cos(B/2)$.\nExpanding $\sin B = 2 \sin(B/2) \cos(B/2)$ yields $4 \sin(B/2) \cos(B/2) = \sqrt{2} \cos(B/2)$.\nSince $B \neq \pi$, $\cos(B/2) \neq 0$, so $\sin(B/2) = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}}$.\nFrom this, $\cos(B/2) = \sqrt{1 - \frac{1}{8}} = \frac{\sqrt{7}}{2\sqrt{2}}$.\nThen $\cos B = 1 - 2 \sin^2(B/2) = 1 - 2(1/8) = 3/4$. **Step 3: Calculate the area of the triangle** The area $a$ (often denoted $\Delta$) is $\Delta = \frac{abc}{4R} = \frac{abc}{4}$.\nWe have $ac = (2 \sin A)(2 \sin C) = 4 \sin A \sin C = 2(\cos(A-C) - \cos(A+C)) = 2(\cos(\pi/2) - \cos(\pi-B)) = 2 \cos B$.\nSubstituting $\cos B = 3/4$, we get $ac = 2(3/4) = 3/2$.\nThe side $b = 2 \sin B = 4 \sin(B/2) \cos(B/2) = 4 \left( \frac{1}{2\sqrt{2}} \right) \left( \frac{\sqrt{7}}{2\sqrt{2}} \right) = \frac{\sqrt{7}}{2}$.\nThus, the area is $a = \frac{ac \cdot b}{4} = \frac{(3/2)(\sqrt{7}/2)}{4} = \frac{3\sqrt{7}}{16}$. **Step 4: Calculate the required expression** We need the value of $(64a)^2$.\n$64a = 64 \left(\frac{3\sqrt{7}}{16}\right) = 12\sqrt{7}$.\n$(64a)^2 = (12\sqrt{7})^2 = 144 \times 7 = 1008$.