Let's determine those three-digit prime numbers in which the product of the digits is 189.
Decompose 189 into prime factors and list its positive divisors: $189=3^{3} \cdot 7$, the divisors are: $1,3,7,9,21,27,63$ and 189. Among the divisors, only the single-digit ones, i.e., 1, 3, 7, 9, can be digits. These must be chosen such that their product is 189.

The digit 1 cannot be used, because then the product of the other two digits would be at most $9 \cdot 9=81$. The digits 9 and 7 can appear at most once, because neither $9 \cdot 9=81$ nor $7 \cdot 7=49$ is a divisor of 189.

Therefore, the digits of the three-digit number are: 3, 7, and 9.

The possible three-digit numbers are: $379, 397, 739, 793, 937, 973$.

Among them, $379, 397, 739, 937$ are prime, so these are the solutions to the problem.

Paulin Dániel (Fazekas M. Fóv. Gyak. Gimn., 8th grade)