According to the same book, if we have 81 coins, one of which is counterfeit and we know it's heavier(or lighter) than the remaining one, then the minimum number of weighings is (4).
Thus, in our problem, we need at least one more weighing to determine whether the counterfeit is heavier(or lighter). We actually need 2 weighing to do this in our case, but we're lucky enough to only need 3 more weighings due to the division of 81 into piles of 27 coins. Therefore, 4 is not possible