Maximizing with integration

The integral will be greatest when $E$ is the largest region for which the integrand is non-negative, namely the ellipsoid $\hat{E}\;:\; x^2 + 2y^2 + 3z^2 \le 1$ . The change of variables $\xi = x$ , $\eta = \sqrt{2}y$ , $\zeta =\sqrt{3}z$ makes this greatest integral equal to $\begin{aligned} \iiint_{\hat{E}} (1 - x^2 - 2y^2 - 3z^2)\,dx\,dy\,dz & = \; \tfrac{1}{\sqrt{6}}\iiint_{\xi^2 + \eta^2 + \zeta^2 \le 1}(1-\xi^2-\eta^2-\zeta^2)\,d\xi\,d\eta\,d\zeta \; = \; \tfrac{1}{\sqrt{6}}\int_0^1 dr\int_0^\pi d\theta \int_0^{2\pi} (1-r^2)r^2sin\theta \\ & = \; \tfrac{4\pi}{\sqrt{6}}\int_0^1 (r^2 - r^4)\,dr \; = \; \tfrac{4\pi}{\sqrt{6}} \times \tfrac{2}{15} \; = \; \tfrac{4}{15}\pi \sqrt{\tfrac23} \end{aligned}$ making the answer $4 + 15 + 2 + 3 = \boxed{24}$ .