This question is extremely close to this one

Covering the primes by 3-term APs ?

though not exactly the same.

For much the same reasons as described in the answer given there, the answer to your question is almost certainly yes, but a proof is beyond current technology, exactly as you suggest. I'm not aware that the problem has a specific name.

To show that 3 belongs to a 3PAP is of course trivial: it belongs to 3,5,7 or 3,7,11. Showing that there are infinitely many such 3PAPs is, as you point out, a problem of the same level as difficulty as the Sophie Germain primes conjecture or the twin primes conjecture.

For a general p, I find it extremely unlikely that you could show that there is a k > 0 such that p + k, p + 2k are both prime without showing that there are infinitely many. Proving this for even one value of p would be a huge advance.

I think you could show that almost all primes p do have this property using the Hardy-Littlewood circle method.

This question is extremely close to this one

Covering the primes by 3-term APs ?

though not exactly the same.

For much the same reasons as described in the answer given there, the answer to your question is almost certainly yes, but a proof is beyond current technology, exactly as you suggest. I'm not aware that the problem has a specific name.

To show that 3 belongs to a 3PAP is of course trivial: it belongs to 3,5,7 or 3,7,11. Showing that there are infinitely many such 3PAPs is, as you point out, a problem of the same level as difficulty as the Sophie Germain primes conjecture or the twin primes conjecture.

For a general p, I find it extremely unlikely that you could show that there is a k > 0 such that p + k, p + 2k are both prime without showing that there are infinitely many. Proving this for even one value of p would be a huge advance.

I think you could show that almost all primes p do have this property using the Hardy-Littlewood circle method.