I am trying to gain some intuition for telling when a variety has quotient singularities.
The example I am focusing on here is the affine variety $X$ which is cut out by the equation $ux+vy+wz=0$ in $\mathbb C^6=\mathbb C^3\times\mathbb C^3$, where $u,v,w$ are the first three coordinates and $x,y,z$ are the last three coordinates. In this case, the only singularity is $0\in X$.
Is $0\in X$ a quotient singularity?
I know that for toric varieties, "$\mathbb Q$-factorial" and "(at worst) quotient singularities" are equivalent, but this variety $X$ is not toric. In general, I believe that it is true that "(at worst) quotient singularities" implies "$\mathbb Q$-factorial" (right?). This variety $X$ is factorial, so the contrapositive of the previous sentence does not apply here.
Note that any punctured neighborhood $U \subset X$ of $0 \in X$ is simply connected, hence $(X,0)$ is not a quotient singularity.
Indeed, it is homotopy equivalent to an $S^1$-bundle over the flag variety $$ F = \{ ux+vy+wz = 0 \} \subset \mathbb{P}^2 \times \mathbb{P}^2, $$ such that the associated line bundle over $F$ has bidegree $(1,1)$. Since $F$ is simply connected, the exact sequence of the homotopy groups takes the form $$ \dots \to \pi_2(F) \to \pi_1(S^1) \to \pi_1(U) \to 0 $$ and the map $\mathbb{Z}^2 = \pi_2(F) \to \pi_1(S^1) = \mathbb{Z}$ in it is surjective, hence the claim.