The sum of any three of them must be an odd prime because $2<1+2+3$. In addition, it mustn't equal to $3$.
If there're $1$ or $2$ even numbers, select one of them as well as any other two odd numbers. Now the sum is even.
If there're no less than $3$ even numbers, select any three of them and the sum is even.
Therefore, all $5$ numbers are odd.
Let $S_i=\{x\mid x\equiv i\pmod3\}(i=0,1,2)$.
There cannot be $3$ numbers from the same set or else their sum is a multiple of $3$.
Therefore, all $5$ numbers come from at least $\left[\frac{5-1}{3-1}\right]+1=3$ sets. (Pigeon theorem)
Now select one from each set and their sum $\equiv1+2+3\equiv0\pmod3$. It's not prime.
So, it isn't possible.