it would be a better puzzle if properly stated. the balls must be identifiable, not identical. identifiability is what differentiates this problem from the 8- or 9-ball, two-weighings problem.

NOTATION
L: heavier on the left
R: heavier on the right
B: balanced
h: heavier ball
l: lighter ball
thus,
1L: first use of balance weighted to the left
3h: third ball might be heavier
8hl: eighth ball might be heavier or lighter

) possible: {1hl,2hl,3hl,4hl,5hl,6hl,7hl,8hl,9hl,10hl,11hl,12hl}
1) balance ^12345678
1L) possible: {1h,2h,3h,4h,5l,6l,7l,8l}
1L.2) balance ¹²⁵³⁴⁶
1L.2L) possible: {1h,2h,6l}
1L.2L.3) balance ¹²
1L.2L.3L) *1h
1L.2L.3R) *2h
1L.2L.3B) *6l
1L.2R) possible: {3h,4h,5l}
1L.2R.3) balance ³⁴
1L.2R.3L) *3h
1L.2R.3R) *4h
1L.2R.3B) *5l
1L.2B) possible: {7l,8l}
1L.2B.3) balance ⁷⁸
1L.2B.3L) *8l
1L.2B.3R) *7l
1R) possible: {1l,2l,3l,4l,5h,6h,7h,8h}
//same as 1L, just swap ¹²³⁴<=>⁵⁶⁷⁸
1B) possible: {9hl,10hl,11hl,12hl}
1B.2) balance [9,10,11][1,2,3]
1B.2L) possible: {9h,10h,11h}
1B.2L.3) balance ⁹¹⁰
1B.2L.3L) *9h
1B.2L.3R) *10h
1B.2L.3B) *11h
1B.2R) possible: {9l,10l,11l}
1B.2R.3) balance ⁹¹⁰
1B.2R.3L) *10l
1B.2R.3R) *9l
1B.2R.3B) *11l
1B.2B) possible: {12hl}
1B.2B.3) balance ¹²+¹
1B.2B.3L) *12h
1B.2B.3R) *12l

[[ DONE ]]

But, just for kicks, a different solution to 1B.2:
1B) possible: {9hl,10hl,11hl,12hl}
1B.2) balance [1,9]+[10,11]
1B.2*L) possible: {9h,10l,11l}
1B.2*L.3) balance ¹⁰¹¹
1B.2*L.3L) 11l
1B.2L.3R) 10l
1B.2L.3B) 9h
1B.2R) possible: {9l,10h,11h}
1B.2*R.3) balance ¹⁰¹¹
1B.2*R.3L) 10h
1B.2R.3R) 11h
1B.2R.3B) 9l
1B.2B) possible: {12hl}
1B.2*B.3) balance ¹²+¹
1B.2*B.3L) 12h
1B.2B.3R) *12l

There are two typing errors in the text above:
• “If 1, 2, and 5 are heavier than 6,7, and 12”
should be 3 instead of 6
o “If they balance, it’s the third light ball (5)”
should be If they balance its the potentially light ball on the other side (7)

woojosh, I thought that also, but I then realized that the problem is that you do not know whether the ball is heavier or lighter.
“One of them is slightly different in weight”

Wouldn’t it be easier to weigh six balls against the other six…one side would contain the heavier ball, and would be the heavy side. Remove the other six and weigh the remaining six in groups of three…then like before, one side would be heavier, leaving you with three balls. Remove 1, if the scales are balanced, you are holding the heavy ball…if one side of the scale is heavy, that is the heavy ball…three weighs, relatively simple.