Does such a prime even exist?

Since they are 3 consecutive positive integers, one of $p-1,p,p+1$ must be divisible by 3. The number which is divisible by 3 will only be prime if it is 3 itself. Therefore the only possible solutions can be when either $\dfrac{p+1}{2} = 3$ or $\dfrac{p-1}{4} = 3$ or $p = 3$ . The first case implies that $p =5$ but this would mean $\dfrac{p-1}{4} = 1$ which is not prime. The second case implies that $p = 13$ , which does in fact yield a solution. The final case implies $\dfrac{p-1}{4} = \dfrac{1}{2}$ which is most definitely not a prime. Therefore the sole solution (and thus sum of all solutions) is $\fbox{13}$

Let us suppose that $\dfrac{p-1}{4} > 3$ . Then either $\dfrac{p-1}{4} = 3k+1$ or $\dfrac{p-1}{4} = 3k+2$ for some $k\geq1$ k being a natural number.

Case 1: $\dfrac{p-1}{4} = 3k+1$

Then we have $p = 12k+4+1 = 12k+5$ .But then $\dfrac{p+1}{2} = \dfrac{12k+6}{2} = 6k+3 = 3(2k+1)$ . Clearly $3k+1>1$ because of our definition of k which is atleast 1. Thus $\dfrac{p+1}{2}$ is divisible by 3 and hence not a prime.

Case 2: $\dfrac{p-1}{4} = 3k+2$

Then $p = 12k + 9 = 3(4k+3)$ . Again, $p$ is divisible by 3 as $4k+3 > 1$ and hence $p$ is not a prime.

Thus $p$ is neither of the form $3k+1$ or $3k+2$ . Therefore $\dfrac{p-1}{4}$ is not greater than $3$ . This leaves us $2$ choices of $\dfrac{p-1}{4}$ , namely, $\dfrac{p-1}{4}=2$ or $\dfrac{p-1}{4} = 3$ . But $\dfrac{p-1}{4} = 2$ implies $p=9$ which is not a prime. Lastly $\dfrac{p-1}{4} = 3$ implies $p = 13$ . Indeed, $\dfrac{p+1}{2} = 7$ which is also a prime. Thus $p=13$ is the only solution and hence the answer is $\boxed{13}$ .