When we have three orbit case, we would specifically have the subcase such that $r_1,r_2,r_3=2,3,3$ where each $r_i$ represents the order of a stabilizer of some pole $p_i$.
If we let the finite subgroup of $SO(3)$ be $G$ where $|G|=N$, then we get $N=12$ by the equation $\sum_{i}(1-\frac{1}{r_i})=2-\frac{2}{N}$.
Also, if we let $n_i$ be the order of an orbit of $p_i$, then $n_1,n_2,n_3=6,4,4$ by the orbit-stabilizer Thm.
Observing the third orbit whose order is $n_3=4$, we conclude $G$ as a Tetrahedral group.
However, I have a little confusion in this part.
More specifically, when observing the third orbit (say $V$), we first pick a pole (say $z$) in the orbit, then since $r_3=3$, a stabilizer (say $H$) would be a cyclic group generated by a rotation s.t $z$ is a pole of the rotation, and the rotating angle is $\frac{2\pi}{3}$.
So, choosing another pole $u \neq z$ in $V$ would give us vertices (other than $z$) of a tetrahedron by $H$.
But, I am wondering what happens if we picked another pole $x \in V$ s.t $u \neq x \neq z$. (This is possible because $n_3=4$)
Do we get the same tetrahedron as well?
In other words, if $g \in H$, then do $x,g(x),g^2(x),z$ forms the same tetrahedron as the previous case? (i.e. The tetrahedron whose vertices are $u,g(u),g^2(u),z$).
If anything is vague here, please let me know!
Any enlightenment would be appreciated a lot!