How can we find the prime numbers $p, q, r$ such that $5pqr-2p-10r = 270$?
By using the fact that $$5pqr-2p-10r -270\equiv -2p\equiv 0\pmod{5}$$
$$p\equiv 0\pmod{5}$$
and that $p$ is a prime number, we impose that $p = 5$.
Similarly,
$$5pqr-2p-10r -270\equiv 25qr\equiv 0\pmod{2}$$
$$qr\equiv 0\pmod{2}$$
then either $q\equiv 0\pmod{2}$ or $r\equiv 0\pmod{2}$, and so, either $q = 2$ or $r = 2$.
Once you figure out that $p=5$, why not substitute it? $$5pqr-2p-10r=270$$ $$25qr-10-10r=270$$ $$25qr-10r=280$$ $$(5q-2)r=56$$
So we have r is a prime factor of $56$. This leaves 2 possibilities: $r=2$ or $r=7$. $r=7$ is the only case that results in $q$ being prime.
$5pqr-2p=10(r+27)$ so 10 divides $p(5qr-2)$. If $p=5$ we get $5qr-2$ even so, as $q$ and $r$ are prime, one at least is 2 : supposing $r=2$ leads then to $50q-10=290$ thus $q=6$ which is not prime. Hence necessarily $q=2$ thus $50r-10=10r+270$ so $r=7$. We have one solution $(p,q,r)$ with $p=5$ which is $(5,2,7)$. If $p\neq 5,2$ then $GCD(10,p)=1$ and 10 divides $p(5qr-2)$ entails by the Gauss lemma that 10 divides $5qr-2$ so again $q$ and/or $r$ is even and $5qr$ is a multiple of 10 so 10 must divide 2 which is absurd. If $p=2$ then 5 divides $(5qr-2)$ hence 5 divides 2 : absurd again. So the previous solution seems to be the only one.
