Parameter in normal form NE $$\begin{aligned}
&amp;\begin{array}{|c|c|c|c|}
\hline \text{Player 1 / 2}&amp; x &amp; y &amp; z \\
\hline X &amp; a, a &amp; a, 0 &amp; a, 0 \\
\hline Y &amp; 0, a &amp; 3,0 &amp; 0,3 \\
\hline Z &amp; 0, a &amp; 0,3 &amp; 3,0 \\
\hline
\end{array}
\end{aligned}$$
Consider that $a \in (0, +\infty)$ I know that if $a=1$ then the NE is $(X,x)$ yet I don't know how to solve for the NE when $a \in (1,3)$ and $a \in(0,1)$ . My attempt was writting the conditions of indifference to find the mixed strategy NE but somehow I still don't arrive to a clear answer. Also I don't now if I should consider all of the possibilities of the support or if there is any shortcut.
Take any $a\in(0,3)$.
Since $a&gt;0$, X is a best response (BR) to x. Since $a&lt;3$, Y is BR against y, and Z is BR against z.
Similarly, x is BR against X, z against Y, and y against Z.
Hence, your only pure strategy NE is (X,x). The latter is also true if $a=3$.
If $a&gt;3$, X and x are strictly dominant strategies such that (X,x) is the unique NE.
Now back to case $a\in(0,3)$.
Suppose 1 randomizes with probabilities $p_X,p_Y,p_Z$, and 2 randomizes with $q_x,q_y,q_z$. If all those probabilities are positive, both must be indifferent over all three options. Let $\pi^i_a$ be the expected payoff of player $i$ playing action $a$.
$\pi^2_x=a, \pi^2_y=3p_Z, \pi^2_z=3p_Y$ and the $\pi^1$ look similar. You need $\pi^2_x=a = \pi^2_y=3p_Z = \pi^2_z=3p_Y$ if 2 shall randomize over all three. Hence, we need $p_Y=p_Z=\frac{a}{3}$, which is a probability as $a&lt;3$. But since $p_X+p_Y+p_Z=1$ we also need $a\leq \frac{3}{2}$. From the same exercise for player 1, we get $q_y=q_z=\frac{a}{3}$ and also condition $a\leq \frac{3}{2}$.
Hence, if $a&lt;\frac{3}{2}$, we have a mixed strategy Nash equilibrium in which both players randomize over all three of their actions with probabilities $q_y=q_z=pY=p_Z=\frac{a}{3}$ and $q_z=p_Z=1-2\frac{a}{3}=\frac{3-2a}{3}$. Such that both of them expect equilibrium payoff $a$. Any unilateral deviation would also lead to payoff $a$.
We also have an equilibrium in which both players only randomize over $Y,Z$ and $y,z$, respectively, all with probability $\frac{1}{2}$. In this case, equilibrium utility of $\frac{3}{2}$ cannot be improved upon by a deviation to $X$ or $x$, respectively, as $a&lt;\frac{3}{2}$, and the other two pure strategies yield the equilibrium utility by construction.
Both these equilibria would not exist if $a&gt;\frac{3}{2}$, as a deviation to  $X$ or $x$, respectively, would be better or the indifference condition could not be fulfilled.
For any $a&gt;0$, there cannot be an NE in which 1 mixes between X and Y and 2 between x and y, because 2 can only be indifferent between x and y if 1 also plays Z, $p_Z=0 \Rightarrow \pi^2_x=a&gt;\pi^2_y=0$.
Similarly, the other combinations (such as X and Z and x and z) would violate the indifference condition.