Find $P + Q + R$ I was doing questions from previous year's exam paper and I'm stuck on this question.

Suppose $P, Q, R$ are positive integers such that $$PQR + PQ + QR + RP + P + Q + R = 1000$$ Find $P + Q + R$.

It seems easy but I am not getting the point from where I should start. Thank you.
 A: 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.
A: 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.
