$$RRR=111 \times R$$
$$PQPQ=1000 \times P+100 \times Q +10\times P +Q=101 \times (10\times P+Q)$$
so $PQPQ\times RRR=111 \times 101 \times R\times(10 \times P+Q)$.
And we have $639027=PQPQ\times RRR=111 \times 101 \times R\times(10 \times P+Q)$ so we can divide both sides by $111\times101$ to give $$57=R\times(10\times P+Q)\;.$$
The only factors of $57$ are $1,3,19$ and $57$.
$R$ must divide $57$ and because $R$ must be a single digit number it can either be $1$ or $3$.
If $R$ is $3$ then $10\times P+Q=19$ so $P=1$ and $Q=9$.
If $R$ is $1$ then $10\times P+Q=57$ so $P=5$ and $Q=7$.
So there are two solutions (check that these both work) and in both cases $P+Q+R=13$.
This problem is taken from the UKMT Mathematical Challenges.