Use more than one for in the same list comprehension:

[(i, j, k) 
 for i in (11, 22, 33) 
 for j in (44, 55, 66) 
 for k in (77, 88, 99)]

This results in all possible permutations (a cartesian product), where enumeration starts on the right:

[(11, 44, 77), (11, 44, 88), (11, 44, 99), 
 (11, 55, 77), (11, 55, 88), (11, 55, 99), 
 (11, 66, 77), (11, 66, 88), (11, 66, 99), 

 (22, 44, 77), (22, 44, 88), (22, 44, 99), 
 (22, 55, 77), (22, 55, 88), (22, 55, 99), 
 (22, 66, 77), (22, 66, 88), (22, 66, 99), 

 (33, 44, 77), (33, 44, 88), (33, 44, 99), 
 (33, 55, 77), (33, 55, 88), (33, 55, 99), 
 (33, 66, 77), (33, 66, 88), (33, 66, 99)]