Question 15
Tetrahedron $A B C D$ has side lengths $A B=6$, $B D=6 \sqrt{2}$, $B C=10$, $A C=8$, $C D=10$, and $A D=6$. The distance from vertex $A$ to face $B C D$ can be written as $\frac{a \sqrt{b}}{c}$, where $a, b, c$ are positive integers, $b$ is square-free, and $\operatorname{gcd}(a, c)=1$. Find $100 a+10 b+c$.