Answer: $a=2, b=3, c=5$ and all sets obtained by permuting these numbers.

Solution. Suppose that $a<b<c$, then $a b<b c$ and $a c<b c$.

If $a \geqslant 3$, then $a b+b c+c a<3 b c \leqslant a b c$, which is a contradiction. The number $a$ is prime, so $a=2$, and then we get $2 b+2 c+b c \geqslant 2 b c$, or $0 \geqslant b c-2 b-2 c$. Adding 4 to both sides, we can write the last inequality as:

$$
4 \geqslant(b-2)(c-2)
$$

Since $3 \leqslant b<c$ and each factor on the right side does not exceed 4, the inequality is satisfied only for the numbers $b=3, c=5$.

Criteria. Only the set $(2,3,5)$ or its permutations are indicated -1 point. Proved that the smallest number is 2 - another 3 points. No points are deducted for the absence of permutations of the set $(2,3,5)$. Full solution -7 points.
3. Find all distinct prime numbers $a, b$ and $c$ such that $a b+b c+c a \geqslant a b c$.