jmc

We have $${\left(a\sqrt{2}+b\sqrt{3}+c\sqrt{5}\right)}^2=104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006,$$or $$2ab\sqrt{6}+2ac\sqrt{10}+2bc\sqrt{15}+(2a^2+3b^2+5c^2)=104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006.$$Since $a,b,c$ are integers, we get the four equations $$\begin{array}{rl}2ab& =104,\\ 2ac& =468,\\ 2bc& =144,\\ 2a^2+3b^2+5c^2& =2006.\end{array}$$Thus, $ab=52$, $ac=234$, and $bc=72$. To find $abc$, we multiply these three equations together, getting $$(abc{)}^2=52\cdot 234\cdot 72=(2^2\cdot 13)\cdot (2\cdot 3^2\cdot 13)\cdot (2^3\cdot 3^2)=2^6\cdot 3^4\cdot 13^2.$$Then, $abc=2^3\cdot 3^2\cdot 13=\boxed{936}$.

We can solve the equations $ab=52,$ $ac=234,$ and $bc=72$ to get $a=13,$ $b=4,$ and $c=18,$ which do satisfy $2a^2+3b^2+5c^2=2006.$ Thus, such positive integers $a,$ $b,$ $c$ do exist.