Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

We basically want to sum them, we can place either plus or minus in front of them or make an element disappear, but we can't make all of them disappear. Instead of considering three numbers, consider the set ${\pm a_1,\pm a_2,\pm a_3}$

Case 1:

If one of them is 0 $(mod 7)$, then we can make the rest of the numbers disappear and that will be enough.

Case 2:

If two of them are equal we cancel them and make the last one dissapear

Case 3:

The last case is that all of the six numbers in the set is distinct $(mod 7)$. It is easy to see that the 3 numbers have to be $\pm1, \pm2, \pm3$. So we arrange them to be.

\[1+2-3 = 0 \ (mod \ 7)\]

So we are done.