Find $P + Q + R$

Note that

$(1 + P) (1 + Q) (1 + R) = 1 + P + Q + R$ $+ PQ+ PR + QR + PQR = 1 + 1000 = 1001, \tag{1}$

by the hypothesis on $P$, $Q$, $R$; also,

$1001 = 7 \cdot 11 \cdot 13, \tag{2}$

all primes; thus we may take

$P = 6; \;\; R = 10; \;\; Q = 12, \tag{3}$

or some permutation thereof; in any event, we have

$P + Q + R = 28. \tag{4}$

Note Added in Edit, Saturday 12 August 2017 10:04 PM PST: The solution is unique up to a permutation of $(P, Q, R) = (6, 10 ,12)$ by virtue of the fact that $1001$ may be factored into three non-unit positive integers in exactly one way, $1001 = 7 \cdot 11 \cdot 13$, since $7$, $11$, and $13$ are all primes. This follows from the Fundamental Theorem of Arithmetic. These remarks added at the suggestion of user21820 made 5 May 2015. Sorry about the delayed response. End of Note.

If you allow some $p,q,r$ to be $0$, you also get $$p+q+r=1000$$ $$p+q+r=88$$ $$p+q+r=100$$ $$p+q+r=148$$

for $$p=1000,q=0,r=0$$ $$p=76,q=12,r=0$$ $$p=90,q=10,r=0$$ $$p=142,q=6,r=0$$

respectively, or any permutation thereof.