3A. Find all natural numbers $n$ for which the numbers

$$
3 n-4,4 n-5,5 n-3
$$

are prime.
Solution. Let's note that $3n-4+4n-5+5n-3=12n-12$ is even. Therefore, at least one of the numbers $3n-4, 4n-5, 5n-3$ must be even. Since the only even prime number is 2, at least one of them must be 2. Clearly, $4n-5$ cannot be 2. So it must be $3n-4$ or $5n-3$. If $3n-4=2$, then $n=2$, and the numbers would be 2, 3, and 7, which are all prime numbers. If $5n-3=2$, then $n=1$, and the numbers would be $-1, -1$, and 2. But these are not all prime. Therefore, only for $n=2$ are the numbers $3n-4, 4n-5, 5n-3$ prime.