Noticing that the left-hand sides are symmetric in $ x $, $ y $, and $ z $ (in that any relabelling of the variables leads to one of the given left-hand sides), we sum all three equations to obtain $ x+y+z=40 $. Subtracting the first equation from $ x+y+z=40 $ gives $ 2z=48\implies z=24 $. Similarly, subtracting the second equation from $ x+y+z=40 $ gives $ y=11 $. Finally, subtracting the third equation from $ x+y+z=40 $ gives $ x=5 $ so that $ xyz=(5)(11)(24)=\boxed{1320} $.