jmc

Denote the height of $△ACD$ as $h$, $x=AD=CD$, and $y=BD$. Using the Pythagorean theorem, we find that $h^2=y^2-6^2$ and $h^2=x^2-15^2$. Thus, $y^2-36=x^2-225⟹x^2-y^2=189$. The LHS is difference of squares, so $(x+y)(x-y)=189$. As both $x,y$ are integers, $x+y,x-y$ must be integral divisors of $189$. The pairs of divisors of $189$ are $(1,189)(3,63)(7,27)(9,21)$. This yields the four potential sets for $(x,y)$ as $(95,94)(33,30)(17,10)(15,6)$. The last is not a possibility since it simply degenerates into a line. The sum of the three possible perimeters of $△ACD$ is equal to $3(AC)+2(x_1+x_2+x_3)=90+2(95+33+17)=\boxed{380}$.