Let $O$ be the cirumcenter of triangle $A B D$. By the inscribed angle theorem, $\angle A O C=90^{\circ}$ and $\angle B O C=60^{\circ}$. Let $A O=B O=C O=x$ and $C O=y$. By the Pythagorean theorem on triangle $A O C$, $$x^{2}+y^{2}=49$$ and by the Law of Cosines on triangle $B O C$, $$x^{2}-x y+y^{2}=25$$ It suffices to find the sum of all possible values of $B D=\sqrt{3} x$. Since the two conditions on $x$ and $y$ are both symmetric, the answer is equal to $$\sqrt{3}(x+y)=\sqrt{9\left(x^{2}+y^{2}\right)-6\left(x^{2}-x y+y^{2}\right)}=\sqrt{291} .$$ It is easy to check that both solutions generate valid configurations.