Given a board (divided into squares) of the size: a) \(10\times 12\), b) \(9\times 10\), c) \(9\times 11\), consider the game with two players where: in one turn a player is allowed to cross out any row or any column if there is at least one square not crossed out. The loser is the one who cannot make a move. Is there a winning strategy for one of the players?
There is a system of equations \[\begin{aligned} * x + * y + * z &= 0,\\ * x + * y + * z &= 0,\\ * x + * y + * z &= 0. \end{aligned}\] Two people alternately enter a number instead of a star. Prove that the player that goes first can always ensure that the system has a non-zero solution.