The Defective Ball

#Why three weighings can suffice

Each weighing returns one of three outcomes, left heavier, balanced, or right heavier, so three weighings yield at most $3^3 = 27$ distinguishable results. The answer must name one of $12$ balls together with one of two directions, which is $12 \cdot 2 = 24$ possibilities. Since $27 \ge 24$, the information is available, and the work is to design weighings that extract it.

#First weighing

Weigh $\{1,2,3,4\}$ against $\{5,6,7,8\}$ and keep $\{9,10,11,12\}$ aside.

#The balanced branch

If the pans match, the odd ball lies in $\{9,10,11,12\}$ and balls $1$ through $8$ are known good. Weigh $\{9,10,11\}$ against the good balls $\{1,2,3\}$.

If they match, ball $12$ is the odd one, and a final weighing of $12$ against any good ball reveals its direction. If $\{9,10,11\}$ is heavier, the odd ball is among $9,10,11$ and is heavy, so weighing $9$ against $10$ names it through the heavier pan, or fixes $11$ when they match. If $\{9,10,11\}$ is lighter, the same final weighing names the light ball.

#The tilted branch

Suppose instead $\{1,2,3,4\}$ is heavier. Then either one of $1,2,3,4$ is heavy or one of $5,6,7,8$ is light. Weigh $\{1,2,5\}$ against $\{3,4,6\}$.

If they match, the cause sits among the untouched $\{7,8\}$ and must be light, so weighing $7$ against a good ball finds it. If $\{1,2,5\}$ is heavier, the only consistent causes are $1$ or $2$ heavy or $6$ light, settled by weighing $1$ against $2$. If $\{3,4,6\}$ is heavier, the causes are $3$ or $4$ heavy or $5$ light, settled by weighing $3$ against $4$.

The remaining first-weighing outcome, $\{5,6,7,8\}$ heavier, mirrors this branch with the two pans' roles swapped.

#Result

Every outcome path terminates on a single ball with its direction, so three weighings always solve the puzzle.