Finding $3$ distinct prime numbers $a| (bc+b+c)$, $b|(ac+a+c)$, $c|(ab+a+b)$ How to find $3$ prime numbers $a,b$ and $c$ such that:
$$a| (bc+b+c)$$
$$b|(ac+a+c)$$
$$c|(ab+a+b)$$

There are no such primes:
Without loss of generality assume $a<b<c$. Clearly 2 cannot be one of the primes, so $a\geq 3$ , $b\geq 5$ and $c\geq 7$.
Now $abc+ab+bc+ca+a+b+c$ is divisible by $a$, by $b$ and by $c$ so it is also divisible by $abc$, 
but this is impossible since
$1<\frac{abc+ab+bc+ca+a+b+c}{abc}<\frac{a+1}{a}\frac{b+1}{b}\frac{c+1}{c}\leq\frac{4}{3}\frac{6}{5}\frac{8}{7}<2$
