Let's name the coins A, B, C, D, E, and F. So in our first weight we put the coins A, B, C, and D. In the second weight we put C, D, E, and F. So from here there are 2 cases. Case 1: If the both weights that appeared where the same then the conterfeit coin was in both our weighings so the counterfeit coin is either C or D. then in our last weight we put C and another 3 coins we know are not counterfeit (A, B, E, F) so if the weights that appear are the same than in the two last weighings then C is the counterfeit. If the weight that appears is different then D is the counterfeit. Case 2: If the first two weights were different, then we only know that C and D are not counterfeit. So let's say we put A and E and weight them. Lets say that from the first weight and second weight we got X and Y respectively in the scale. And on the third weight we got Z. then if we multiply the weight of Z times 2 and it is equal to one of the first two weights then A and E are not the counterfeit. If Z times 2 is equal to X then F is the counterfeit but if it is equal to Y then B is the counterfeit. In another case if Z times 2 is not equal to neither X nor Y then the counterfeit coin is A or E. If the counterfeit weights less then this last weight is the lighter and the one with all the authentic ones is the heaviest and if the counterfeit is heavier then this last one is the heaviest and the one in which all are authentic is the lightest. So the weight in between is one of the first two and has the counterfeit coin and because in the first two weights A and E were not together then we can distinguish which one is the counterfeit and thats it.

Here's the official solution:http://www.math.toronto.edu/oz/turgor/archives/TT2005F_SOsolutions.pdf please give your other weightings Let the coins be A, B, C, D, E and F. In three weighings, we determine the average weight m of C and E, the average weight n of D and F, and the average weight k of B, E and F. If m = n = k, the fake coin is A. If m = n ̸= k, the fake coin is B. If m ̸= n = k, the fake coin is C. If k = m ̸= n, the fake coin is D. If k ̸= m ̸= n ̸= k, then the fake coin is E or F. This can be distinguished since 2m + n = 3k if it is E, and m + 2n = 3k if it is F.