I'm assuming primes are >= 2 with exactly 2 factors: 2,3,5,7,11...

Since 2 is the only even prime it places tight constraints.  Negative integers allow a lot of wiggle room, though.

If the numbers are all even there is no solution.

If there's one odd and two evens, and these evens sum to 2.  

For example, (0,2,r) where r is the smaller of a twin prime pair.

But we can also have (-2,4,r) where r is not necessarily prime be is two above primes that are 6 apart (sexy primes) for example (-2,4,15)

In this fashion, we can find a triple for any primes that differ by 4n+2

If there's two odds and one even, then the odds must sum to 2.  We can have (1,1,2).

We can also do (-1,3,r) where r is one larger than primes that are 4 apart (-1,3,6) or example.

(-3,5,r) will work for primes that are 8 apart: (-3,5,10)

In this fashion, we can find a triple for any primes that differ by 4n.

So for any trio of primes (2,p,q) there is a trio of integers (a,2-a,c) whose sums are 2, c+a, c+2-a where 2a-2=q-p

I don't know if this is quite the same as determining an expression that generates all solutions.  It seems a little backwards, but since the twin prime conjecture is unsolved, that's the best I can do.