Finding volume of convex polyhedron given vertices I am trying to compute the volume of the convex polyhedron with vertices $(0,0,0)$, $(1,0,0)$, $(0,2,0)$, $(0,0,3)$, and $(10,10,10)$. I am supposed to use a triple integral but am struggling with how to set it up.
 A: First we can find the equation for the plane that goes through $A = (1,0,0)$, $B = (0,2,0)$, and $C = (0,0,3)$. It is: $x + \dfrac{y}{2} + \dfrac{z}{3} = 1$ or $6x + 3y + 2z - 6 = 0$. Now let $O = (0,0,0)$, and $D = (10, 10, 10)$. We now find the volume $V_1$ of the tetrahedron $OABC$. $V_1 = \frac{1}{6}\cdot OA\cdot OB\cdot OC = \dfrac{1\cdot 2\cdot 3}{6} = 1$. Let $d$ = distance from $O$ to the plane $ABC$, and $S$ = area of triangle $ABC$. Then $3V_1 = S\cdot d$, and $d = |\dfrac{6\cdot 0 + 3\cdot 0 + 2\cdot 0 - 6}{\sqrt{6^2 + 3^2 + 2^2}}| = \dfrac{6}{7} $. So $3\cdot 1 = S\cdot \dfrac{6}{7}$. Thus $S = \dfrac{7}{2}$. Next let $k$ be the distance from $D$ to the plane $ABC$. Then $k = \dfrac{6\cdot 10 + 3\cdot 10 + 2\cdot 10 - 6}{7} = \dfrac{104}{7}$. So let $V_2$ be the volume of the tetrahedron $ABCD$. So $V_2 = \dfrac{S\cdot k}{3} = \dfrac{1}{3}\cdot \dfrac{7}{2}\cdot \dfrac{104}{7} = \dfrac{52}{3}$. So let $V$ be the volume of the polyhedra $OABCD$, then $V = V_1 + V_2 = 1 + \dfrac{52}{3} = \dfrac{55}{3}$.
