By opening the brackets and substituting $n$ we get: $$pqr + (p+q+1)r = pqr + 138 $$$$ (p+q+1)r = 138$$The prime factorization of $138$ is: $$138=2\cdot3\cdot23$$So: $$(p+q+1)r = 2\cdot3\cdot23$$p+q+1 can't be two or three because it is greater than both numbers. From now on we just take the cases: Case 1: $r=2$ and $p+q+1=69$ Simply with guessing and calculating $n$ the solutions we get are: $$(p,q,r,n)=(7,61,2,854) ; (61, 7, 2, 854) ; (31, 37, 2, 2294) ; (37, 31, 2, 2294)$$Case 2: $r=3$ and $p+q+1=46$ Solutions: $$(p,q,r,n)=(2, 43, 3, 258) ; (43, 2 ,3, 258)$$Case 3: $r=6$ and $p+q+1=23$ Solutions: $$(p,q,r,n)=(17, 5, 6, 510) ; (5, 17, 6, 510) ; (3, 19, 6, 342) ; (19, 3, 6, 342) ; (9, 13, 6, 702) ; (13, 9, 6, 702) ; (11, 11, 6, 726) $$Case 4: $r=23$ and $p+q+1=6$ Solutions: $$(p,q,r,n) = (2, 3, 23, 138) ; (3, 2, 23, 138)$$And we are done!