Prime Numbers

Notice that their sum is even. This implies that one of them must be 2. Checking the three cases, we see the only possibility for $n$ is $\boxed{2}$

If $n$ is odd, then $5n-3$ is even. If $n$ is even, then $3n-4$ is even. It is therefore not possible that all three numbers are odd.

However, all prime numbers are odd, except for 2. Therefore one of the three numbers must be two. For $n>1$ , the first value is the smallest. Thus the only solution has $3n-4 = 2\,\Longrightarrow\,n=2.$

Notice that if n is odd then $\ 5n-3$ is never prime. To prove this assertion, let n = 2k+1 (as n is odd) to obtain: $\ 5(2k+1) -3 = 10k +2 = 2(5k+1) \rightarrow\ 5k+1 = 1 \Rightarrow\ k = 0$ $\Rightarrow\ n = 1 \ so \ 3n-4 = -1 < 0$ Hence, 2 is the only option and it is indeed a solution.