Let a, b, and c be the three numbers. Then, ab+c, ac+b, bc+a, |ab-c|, |ac-b|, and |bc-a| are all prime. If a, b, and c are all odd, then all six numbers are even. Since they are all prime, they are all 2. Therefore, ab+c=2 and |ab-c|=2. If ab-c=2, then ab=2 and c=0, which is not prime. If ab-c=-2, then c-ab=2, so ab=0 and c=2. Since ab=0, either a=0 or b=0, but 0 is not prime. Therefore, a, b, and c cannot all be odd, so 2 has to be one of the primes.
Let a=2. Then, 2b+c, 2c+b, bc+2, |2b-c|, |2c-b|, and |bc-2| are all prime. Suppose neither b nor c is 3. Then, bc is not divisible by 3, so bc=1 or 2 mod 3. If bc=1 mod 3, then bc+2 is divisible by 3. However, bc+2 is prime, so bc+2=3. Then, bc=1, but no primes can multiply to 1. If bc=2 mod 3, then |bc-2| is divisible by 3, so |bc-2|=3. If bc-2=3, then bc=5, which is not the product of two primes. If bc-2=-3, then bc=-1, which is not the product of two primes. Therefore, either b or c is 3.
Let b=3. Then, 6+c, 2c+3, 3c+2, |6-c|, |2c-3|, and |3c-2| are all prime. Now, c is either 0, 1, 2, 3, 4, 5, or 6 mod 7. If c=0 mod 7, then c=7, but |6-c|=1, which is not prime. If c=1 mod 7, then 6+c=0 mod 7, so 6+c=7. Then, c=1, which is not prime. If c=2 mod 7, then 2c+3=0 mod 7, so 2c+3=7. Then c=2, but 6+c=8, which is not prime. If c=3 mod 7, then |3c-2|=0 mod 7, so |3c-2|=7. If 3c-2=7, then c=3, but 2c+3=9, which is not prime. If c=4 mod 7, then 3c+2=0 mod 7, so 3c+2=7. However, c=5/3, which is not prime. If c=5 mod 7, then |2c-3|=0 mod 7, so |2c-3|=7. If 2c-3=7, then c=5, but |6-c|=1, which is not prime. If 2c-3=-7, then c=-2, which is not prime. If c=6 mod 7, then |6-c|=0 mod 7, so |6-c|=7. If 6-c=7, then c=-1, which is not prime. If 6-c=-7, then c-6=7, so c=13. The problem shows that this works. Therefore, (2, 3, 13) is the only triplet that works.