It's hard to believe this classic puzzle isn't already documented in MSE, but FWIW, here's one of the standard "static" solutions. "Static" means that the weighings are fixed in advance - we know which coin (or ball, or whatever) goes on which arm of the balance at each step regardless of the outcomes of previous weighings. This seems like a harder problem to solve, but in fact, it makes the process of finding a solution somewhat easier than the more natural case-by-case analysis.
Create a 3-by-12 matrix. The rows correspond to weighings, the columns correspond to coins. Each entry in the matrix is either L (the coin goes on the left of the scale in this weighing), R (the coin goes on the right) or O (the coin is left out in this weighing).
Now if a coin is heavy, it will make all the weighings in which it has L to be left-heavy, all those in which it is R to be right-heavy, and all the ones where it's O to be balanced. For a light coin it's the other way around. So each pair of (coin, weight bias) creates a sequence of results, and you just need to make sure that all those sequences are different for all the (coin, bias) combinations. In other words, the columns of the matrix must all be different, and moreover, none of them should be inverses of each other (since if coin x is the inverse of coin y, you can't tell a heavy coin x from a light coin y).
Here's an example solution (I use +, - and blanks here instead of R, L and O since it's easier to grasp I think) which has a little bit of structure making it easy to verify the requirements.
1 2 3 4 5 6 7 8 9 A B C
+ + - + - - + -
+ + + - - - - +
+ - + + - - - +
EDIT: to explain once again what all this means, in the first weighing we pit coins 1, 4, 6 and 10 against coins 5, 7, 9 and 12. The second weighing has 2, 4, 5, 11 against 6, 7, 8, 10 and the last one has 3, 5, 6, 12 against 4, 8, 9, 11. So the weighings are always 4 against 4, and as mentioned, the result of weighing #1 don't matter at all for how we choose to handle the other weighings. It's all static.
If coin 1 is heavy, for example, we'll get "Left, balanced, balanced". No other coin can do that by being heavy (there's no other column like this one) and no other coin can do that by being light (since there's no column that goes "- blank blank").