In solving $2$-player games (be it $2\times2$ or $3\times3$ or $n\times n$), we fix each of player 1's pure strategy, and check player 2's best response (BR) to that strategy, and then do the same by reversing the player roles (fix each of player 2's pure strategy, and check player 1's BR to it).
With $3$-player games, the idea is similar: In checking a player's BRs, we fix the other two players' strategies.
Suppose player 1 chooses rows (A, B, C), player 2 chooses columns (D, E, F), and player 3 chooses matrices (G, H, I).
- In checking player 1's BR to (D,G), we focus on the first column (D) of the left matrix (G). Then, inspecting the first element in the payoffs, we see that B gives the highest payoff (3).
- Similarly, player 1's BR to (E,G) is B again.
- From here we can see that player 1's BRs can be found by inspecting the first element of the payoff vectors in each column of each matrix.
- Likewise, player 2's BR to (A,G) is found by comparing the second element in the payoff vector in the first row (A) of the first matrix (G). Since E attains the highest payoff (2) compared to D (with payoff 1) and F (with payoff 1), so E is a BR to (A,G).
- Thus player 2's BRs can be found by inspecting the second element of the payoff vectors in each row of each matrix.
- To see player 3's BR to (A,D), we look at the third number in the top-left cell of each matrix. Playing G gets payoff 3, H gets payoff 4, I gets payoff 2. Hence H is a BR to (A,D).
- Generalizing, player 3's BRs can be found by inspecting the third element of the payoff vectors in the corresponding cells in each matrix.
This procedure shows that (B,D,G) and (B,F,G) are two pure NEs of this game.