Direct Computation

The base is an ellipse with semimajor axis $A = \sqrt{42}$ and semiminor axis $B = \sqrt{26}$ , and these two axes point in the direction of the orthogonal unit vectors $\mathbf{u} = \frac{1}{\sqrt{42}}\left(\begin{array}{c} -5 \\ 4 \\ -1 \end{array}\right) \hspace{2cm} \mathbf{v} \; = \; \frac{1}{\sqrt{26}}\left(\begin{array}{c} 3 \\ 4 \\ 1 \end{array}\right)$ A unit vector normal to the base is thus $\mathbf{w} = \mathbf{u} \times \mathbf{v} \; =\; \frac{1}{\sqrt{243}}\left(\begin{array}{c} 4 \\ 1 \\ -16\end{array}\right)$ Since $\mathbf{w} \cdot \left[\left(\begin{array}{c} 1 \\ 2 \\ 10 \end{array}\right) - \left(\begin{array}{c} 5 \\ 6 \\ -10 \end{array}\right) \right] \; =\; - \frac{340}{\sqrt{243}}$ we deduce that the vertical height of the cone is $H = \tfrac{340}{\sqrt{243}}$ , and hence the volume of the cone is $\frac13 \times \pi AB \times H \; = \; \frac{680}{3}\pi$ making the answer $680+3=\boxed{683}$ .