Let $p=q+r=s-t$ where $r<q$. Obviously $p>2$. If both $q,r>3$ then $p$ is even and greater than $2$, so composite. Thus, $r=2$ and similarly from $s=p+t$, since $p>2$, we have $t=2$. Therefore, $p,p-2$ and $p+2$ are all primes. Consider them modulo $3$. They form a residue class of $3$. Since $p-2$ is at least $3$, we can't have $p=3$. Also, we can't have $p+2=3$ either. Thus, $p-2=3$ and then $p=5,p+2=7$ are all primes.

If both the Prime numbers are odd then their sum & difference are even so both cannot be prime. So, one of the Prime numbers is even that is one of them is 2. So, let p-2, p & p+2 are primes. If p =3k+1, then 3k+2 is a multiple of 3. If p= 3k+2, then p-2 is a multiple of 3. So, p-2 =3 is the only possibility. So, the required prime numbers are 2 & 5 as p =5.