First, we see that faces $A B D, A B C$, and $A C D$ are all right triangles. Now, $A B D$ can be visualized as the base, and it can be seen that side $A C$ is then the height of the tetrahedron, as $A C$ should be perpendicular to both $A B$ and $A D$. Therefore, the area of the base is $\frac{6^{2}}{2}=18$ and the volume of the tetrahedron is $\frac{18 \cdot 8}{3}=48$. Now, let the height to $B C D$ be $h$. The area of triangle $B C D$ comes out to $\frac{1}{2} \cdot 6 \sqrt{2} \cdot \sqrt{82}=6 \sqrt{41}$. This means that the volume is $$48=\frac{6 h \sqrt{41}}{3}=2 h \sqrt{41} \Longrightarrow h=\frac{24}{\sqrt{41}}=\frac{24 \sqrt{41}}{41}$$