The answer is $(13, 5, 7)$. The left side can be rewritten as $36((yz+1)x+z)$. Because $gcd(36, 475)=1$ and $gcd((yz+1)x+z, yz+1) = 1$, we conclude $yz+1 = \pm 36$ and $(yz+1)x + z=\mp 475$.