First of all, if these three primes are distinct, their product would divide $(a^2+1)(b^+1)(c^2+1)$ and hence in particular \[(bc+a)(ca+b)(ab+c) \le (a^2+1)(b^2+1)(c^2+1)\]which, after expanding, is equivalent to \[(abc-1)(a^2+b^2+c^2+1) \le 0\]which is clearly wrong. Hence, two of the primes have to be the same, say $a+bc=b+ac$ i.e. $(a-b)(c-1)=0$. So either $a=b$ or $c=1$. If $a=b$, then $a+bc=a(c+1)$ is prime, i.e. $a=b=1$, but then $c+1$ is prime and divides $4(c^2+1)$, hence divides $8$, hence $c=1$ as well and we get the solution $(1,1,1)$. Else, if all three of $a,b,c$ are distinct, we get that $ab+1$ and $a+b$ are primes $\ge 3$ dividing $2(a^2+1)(b^2+1)$, hence dividing $(a^2+1)(b^2+1)$. But if w.l.o.g. $a<b$, then $ab+1>a^2+1$ and hence $ab+1 \mid b^2+1$ and hence $ab+1 \mid b^2-ab=b(b-a)$, but both factors are positive and smaller than $ab+1$, contradiction. Hence the only solution is $(1,1,1)$.