Actually, I have a different answer. I've also tried solving this problem, and I found out that it is somehow possible at the same time impossible to locate the fake coin with only 3 weighing--there is always a double possible outcome. I suggest that minimum no. of weighing to locate the coin is 4. And here how it goes.......

Divide the coins into 3 groups (4 coins each)

Left side----------------right side

1st weighing:

4 coins---------------4 coin

If the coins weigh even, then the counterfeit coin is in the 3rd group of coins. Otherwise, if the weights are not even, the fake coin is on either side of the scale. We take the coins on either side of the scale, then replace the empty side with the 3rd group of coins.

2nd weighing:

4 coins----------------4 coins

If the group of coins weigh the same, then the fake coin is in the group of coins that was set aside. Otherwise, if they don't weigh the same--the fake coin is in the side of the scale before the actual 2nd weighing took place.

3rd weighing

1 coin---------------1 coin

2 possibilities

1st

If they weigh the same, then we set them aside. So we know that the fake coin is either of the 2 coins that were left.

2nd

If they don't. So we know that the fake is either of the 2 coins.

4th weighing

1 coin---------------1 coin

So know we are down we only 2 coins. To locate the fake coin, we set aside either of the two coins. Then we get any of the ten coins that were set aside in the 1st-3rd weighing. If the coins weigh the same, then the coin that was set aside is the counterfeit. Otherwise, if they don't--then we located the fake coin (if you know what I mean).