Odd ball among twelve in three weighings

Twelve balls look identical, but exactly one is either heavier or lighter than the others, and you do not know which. Using a two-pan balance three times, find the odd ball and say whether it is heavy or light.

Weigh 4 against 4 first. If balanced, the odd ball is among the other four and you have eight good balls: weigh three suspects against three good ones, then resolve with one more weighing. If unbalanced, the eight balls carry a heavy-or-light tag; weigh two heavy suspects plus one light against one heavy, one light and a good ball, so each outcome leaves at most three tagged balls for a final weighing. Since $24\le3^3$ cases, three weighings suffice.