I'll preserve your notation: $M$ is the coadjoint orbit of a regular semisimple element $X \in \mathfrak t^*$ (which you seem to also call $x$).

The orbit $M$ is neither contained in, nor contains, $\mathfrak t^*$. Rather, a conjugate of $X$, which you seem also to be calling $x$, lies in $\mathfrak t^*$ if and only if it is a conjugate by the Weyl group $W = \operatorname N_{\operatorname{SU}(3)}(T)/T$. Thus, since regular elements in this case are strongly regular, $M^T = M \cap \mathfrak t^*$ has order $6$. These are the vertices of your hexagon.

I'll preserve your notation: $M$ is the coadjoint orbit of a regular semisimple element $X \in \mathfrak t^*$ (which you seem to also call $x$). I also assume we're working in characteristic $0$, or at least not $3$.

The orbit $M$ is neither contained in, nor contains, $\mathfrak t^*$. Rather, a conjugate of $X$ lies in $\mathfrak t^*$ if and only if it is a conjugate by the Weyl group $W = \operatorname N_{\operatorname{SU}(3)}(T)/T$. Thus, since regular elements in this case are strongly regular, $M^T = M \cap \mathfrak t^*$ has order $6$. These are the vertices of your hexagon.