Pure Nash Equilibria 3 players game

I'm trying to solve this pure-strategy Nash equilibria of this game below: I highlighted the best pay off for player 1 and 2. But I don't get it when it comes to player 3. The correct answer is (A)

I tried to solve it as a gaming tree. But still I don't get it.

How shall I think?

Three-player games are notoriously tricky to analyze. The ideal way to display them would be a three-dimensional array of cells, each containing three payoffs. But this is difficult to write down on two-dimensional paper. So typically an $$n\times m\times l$$-game is displayed as $$l$$ different $$n\times m$$-matrices. In your example you might think about it in this way:
• Player 1 chooses the row (upper row, $$A$$, or lower row, $$B$$)
• Player 2 chooses the column (left column, $$A$$, or right column, $$B$$)
• Player 3 chooses the matrix (upper matrix, $$A$$, or lower matrix, $$B$$)
Therefore, to highlight the best payoff of player 3, for each of the 4 choices of players 1 and 2 (for each of the 4 cells in the matrices) you have to compare the player-3-payoffs between the upper and the lower matrix. So in the $$(A,A)$$-cell you highlight the $$70$$ in the upper matrix, since it is greater than the $$60$$ in the lower matrix. For the $$(A,B)$$-cell you highlight $$23$$, which is greater than $$0$$, and so on. In the end, the $$(A,A)$$-cell of the upper matrix (corresponding to the choice of $$A$$ by player 3) will have all three payoffs highlighted, and the same will be true for the $$(B,B)$$-cell of the upper matrix (again corresponding to the choice of $$A$$ by player 3). So the pure NE are $$(A,A,A)$$ and $$(B,B,A)$$.