Sorry for my English, I am French but i couldn't find help on the French website (so I am here).
I have a question about this two-player game: $$ \begin{array}{c|cc} & y_1 & y_2 \\ \hline x_1 & a,0 & a,0 \\ x_2 & 1,-1 & -1,1 \\ x_3 & -1,1 & 1,-1 \end{array} $$ When $a<0$ and $a>1$ it's easy because using domination we can find the Nash equilibrium (mixed), but when $0 \leq a \leq 1$ this is more difficult.
If $(p_1, p_2, 1-p_1-p_2)$ is the strategy of the player $1$ and $(q, 1-q)$ the strategy of the player $2$, by the indifference theorem for player $1$ I have $$ E(x_1) = E(x_2) = E(x_3) \iff a = 2q-1 = -2q+1 $$ and for player $2$ I have $$ E(y_1) = E(y_2) \iff -p_2 + (1-p_1-p_2) = p_2 - (1-p_1-p_2) $$ What is the Nash equilibrium (mixed and pure) of this game for $0 \leq a \leq 1$? I can't solve this equation...
First, let's restrict further to $0\lt a\lt1$, since for $a=0$ we can replace any $x_1$ component in the strategy by a corresponding equal mixture of $x_2$ and $x_3$, and for $a=1$ $x_1$ dominates both of the other strategies.
With $a\gt0$, your first equation has no solution, since the right-hand equality implies that $2q-1$, and thus $a$, is $0$. That shows that there's no mixed equilibrium with full support on all of $x_1$, $x_2$ and $x_3$.
From the symmetry of the situation, $q=\frac12$ seems like a sensible candidate for a mixed equilibrium. This results in $x_1$ being strictly preferable to $x_2$ and $x_3$, and conversely, if the row player plays $x_1$, the column player is indifferent between $y_1$ and $y_2$. So the row player playing $x_1$ and the column player playing $y_1$ and $y_2$ both with probability $\frac12$ is a mixed equilibrium. In fact any $q$ with $a\ge\left|2q-1\right|$ yields an equilibrium together with $x_1$.