First, I will test small values of r.
Let r=2. Then p/q-4/3=1, which makes p/q = 7/3. Then (p,q,r)=(7,3,2) is a solution.
Next let r=3. Then p/q-4/4=1. Then p/q=2. But this needs q=1 for p to be prime however 1 is not a prime. So no solution in this case.
Now assume r is a prime of the form 4k+3 (k>=1). Then 4/(r+1) reduces to 1/(k+1). Then p/q = (k+2)/(k+1).
This can only happen when p=3 and q=2. Then k=1 and r=7. Then (p,q,r)=(3,2,7) is a solution.
Finally assume r is a prime of the form 4j+1 (j>=1). Then 4/(r+1) reduces to 2/(2j+1). Then p/q = (2j+3)/(2j+1).
Then p and q are a pair of twin primes. The special twin pair of 5 and 3 occurs when j=1 and r=5. Then (p,q,r)=(5,3,5) is a solution.
For all other pairs of twin primes they are of the form (6h+1,6h-1) for h>=1. Then we need 2j+1=6h-1, or j=3h-1. Then r=4*(3h-1)+1=12h-3=3*(4h-1). But this is composite for all h>=1, so no more solutions in this case.
In summary, the set of solutions is (p,q,r) = (7,3,2), (3,2,7), or (5,3,5).
Solution
