Let the three numbers be $a,b,c$ with $1<a<b<c$. We have $a|bc+1$, $b|ac+1$, and $c|ab+1$. This means $abc|(ab+1)(ac+1)(bc+1)\implies abc|ab+ac+bc+1\implies abc\leq ab+ac+bc+1$. If $a\geq 3$, then $abc\geq 3bc$, but since $ab,ac<bc$, we have $ab+ac+bc+1<3bc$, contradiction. Thus, $a=2$. Now, $2bc\leq bc+2(b+c)+1$. If $b\geq 4$, then $bc+2(b+c)+1\leq bc+2(2c-1)+1<c(b+4)\leq 2bc$, contradiction. Thus, $b=3$, and then $c|ab+1=7\implies c=7$, and we're done.