Triple-Prime

$abc = 23(a+b+c) \implies abc \equiv 0 \mod 23$ . Because the variables are prime numbers, at least one of the numbers must be $23$ . Because of the symmetry of the equation we may assume that $a = 23$ which turns the equation into $bc = 23 + b + c \implies bc - b - c = 23 \implies (b-1)(c-1) = 24$ . So $(b-1)(c-1)$ is one of the positive factorizations of 24 which are $1 \times 24, 2 \times 12, 3 \times 8, 4 \times 6$ . For the variables to be primes, it must be that the numbers in this factorization are one less than a prime and this condition is only satisfied by $2 \times 12, 4 \times 6$ which yield the solutions $(23,3,13),(23,5,7)$ and any permutation of those two.