Rectangular Box Inscribed in Sphere

Let box dimensions be $a, b, c$.

• Surface area $2(ab + bc + ca) = 846$
• Edge length sum $4(a + b + c) = 144 \implies a + b + c = 36$

$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$ $36^2 = a^2 + b^2 + c^2 + 846 \implies a^2 + b^2 + c^2 = 1296 - 846 = 450$

For a box inscribed in a sphere, diagonal $= 2R = \sqrt{a^2 + b^2 + c^2} = \sqrt{450} = 15\sqrt{2}$. Radius of sphere $R = \frac{15\sqrt{2}}{2}$.

$\text{Volume} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \left(\frac{15\sqrt{2}}{2}\right)^3 = \frac{4}{3}\pi \frac{3375 \times 2\sqrt{2}}{8} = 1125\pi\sqrt{2}$