jmc

As shown in the image above, let $D$, $E$, and $F$ be the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Suppose $P$ is the apex of the tetrahedron, and let $O$ be the foot of the altitude from $P$ to $△ABC$. The crux of this problem is the following lemma. Lemma: The point $O$ is the orthocenter of $△ABC$. Proof. Observe that$$O{F}^2-O{E}^2=P{F}^2-P{E}^2=A{F}^2-A{E}^2;$$the first equality follows by the Pythagorean Theorem, while the second follows from $AF=FP$ and $AE=EP$. Thus, by the Perpendicularity Lemma, $AO$ is perpendicular to $FE$ and hence $BC$. Analogously, $O$ lies on the $B$-altitude and $C$-altitude of $△ABC$, and so $O$ is, indeed, the orthocenter of $△ABC$. To find the coordinates of $O$, we need to find the intersection point of altitudes $BE$ and $AD$. The equation of $BE$ is simply $x=16$. $AD$ is perpendicular to line $BC$, so the slope of $AD$ is equal to the negative reciprocal of the slope of $BC$. $BC$ has slope $\frac{24-0}{16-34}=-\frac{4}{3}$, therefore $y=\frac{3}{4}x$. These two lines intersect at $(16,12)$, so that's the base of the height of the tetrahedron. Let $S$ be the foot of altitude $BS$ in $△BPQ$. From the Pythagorean Theorem, $h=\sqrt{B{S}^2-S{O}^2}$. However, since $S$ and $O$ are, by coincidence, the same point, $SO=0$ and $h=12$. The area of the base is $102$, so the volume is $\frac{102\ast 12}{3}=\boxed{408}$.