19 and 97 are primes right?

The equation is equivalent to $19xyz=97(yz-z-1)$ or $19xyz+97=97z\times (y-1).$
$19$ and $97$ are primes and so $97$ should divide one of $x,y,z.$ Also $z$ should divide $97$ which means that $z=97$ is prime.
From this it follows that $y$ divides $98=2\times 49=2\times 7^2;y$ being less than $z=97.$ The possibilities are $y=2,y=7,y=14,y=49.$
If $y=2,19\times 2\times x=2\times 97-98=96$ so that $x$ is not an integer.
If $y=7,19x=83$ so that $x$ is not an integer.
If $y=14,19x=90$ so that $x$ is not an integer.
If $y=49,19\times 49x=49\times 97-98. i.e.,x=5.$
Thus $x=5,y=49,z=97.$
Therefore $x+y+z=5+49+97=\boxed{151}.$