Value of the 3 prime numbers Here's the question

The 3 prime numbers $P,Q,R$ have a value where the product of themselves is 19 times larger than the sum of itself.Find $P^2+Q^2+R^2$.

It means that $19(P+Q+R)=PQR$
Since $P,Q,R$ MUST be an integer,so with the given equation of $P+Q+R=\frac {PQR} {19}$,one of the numbers MUST be 19.From here,I have no idea but just to use Guess and Check.I want a mathematical working other than the guessing style.(Like simultaneous equation or any other working.)
But you can check your answer here.
$P=19$
$Q=3$
$R=11$ (With no order)

As you said, one of them must be $19$. Without loss of generality, take $P=19$. Then, we get,
$$19+Q+R=QR\implies 19+1=1-Q-R+QR=(1-Q)(1-R)\\ \implies (1-Q)(1-R)=20$$
$$20=(\pm 1)\times (\pm 20)=(\pm 2)\times (\pm 10)=(\pm 4)\times (\pm 5)$$
Checking all the possible factor cases yields us that the only possible prime solutions for $Q,R$ is w.l.o.g $Q=3~,~R=11$.
