Find all triplets $(a,b,c)$ of positive integers so that $\gcd(a,b,c)=1$ and $$ \frac{2abc}{(a+b-c)(b+c-a)(c+a-b)} $$ is a positive integer.
What I've done: first I looked with Mathematica for solutions. For $a<150$, $b<1000$ and $c<1000$, I found $(1,1,1)$, $(7,117,121)$ and $(11,39,49)$ as the only valid solutions, which led me to believe these are the only ones. Unfortunately, I have no idea how to prove this assertion. What might be interesting is that these triplets would also be solutions if the problem had stated $abc$ instead of $2abc$, and that in that case the fraction would equal either $1$ or $13$. Now it can be $2$ or $26$. Also, there are more solutions if the fraction is allowed to be negative, but for the moment I'm not interested in those. Any help is appreciated. Thanks.
I am afraid your conjecture is not correct. The situation is quite complex. I have been studying such cubic representation problems for some time.
The basic identity \begin{equation*} \frac{2abc}{(a+b-c)(b+c-a)(c+a-b)}=N \end{equation*} is equivalent to finding points of infinite order on the elliptic curve \begin{equation*} v^2=u^3+2(2N^2-2N-1)u^2+(4N+1)u \end{equation*}
For specifically positive solutions the situation is even more complicated, in that we also need a point where the u-coordinate is negative.
Such situations are quite rare, but other examples come from $N=74, 218$ which give the results in the comment. A novel solution is for $N=250$ with $(97, 10051, 10125)$ as a solution.
We can also use multiples of the points in the elliptic-curve formulation to get larger solutions. For example, $N=26$ has the solution
$a=4739\,2819\,4344\,87$
$b=2634\,1867\,7932\,41$
$c=7228\,3008\,5646\,67$
and we can keep going getting larger and larger solutions.