The task of determining a fake coin of 14

The old, dreadful problem about 12 coins and 3 weighings (only at Habré it was solved here and here ) can be solved in the same way for 13 coins, provided that it is not necessary to determine this counterfeit coin harder or easier (this solution is described in detail , based on the ternary system ).

But I have not seen such a modification yet. The problem is solved for 14 coins, if you can use one (15th) in the weighings as exactly “real”.

The condition of the problem with the modification:
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There are 14 coins. One of them is fake, but it is not known whether it is heavier or lighter than the others. There are also accurate jewelery scales with two cups, which can either be balanced or not, depending on what coins you put on each side.
You must use three and only three weighings. For these three weighings, you need to determine which of the 14 coins is fake.

A heavier or lighter coin is not necessary to determine.
The “correct” 15th coin is given to help you, which you can use in weighings knowing that it is exactly “correct”.