T//W for adjoint type group PGL3 Let $G$ be a reductive algebraic group and $T$ a maximal torus (over $\mathbb{C}$).  It is well known that if $G$ is simply connected type then $T//W = \mathbb{A}^r$.
I want to verify that the following computation is correct (I have some doubts).  I claim that $T//W$ for $G = PGL_3$ is the affine variety with coordinate ring $k[x^2, xy, y^2] = k[A,B,C]/AC - B^2$, i.e. the quadric cone.
The argument is as follows.  The characters of the torus, thought of as monomials, generate the coordinate ring of the torus as a vector space.  For $PGL_3$ the character lattice coincides with the root lattice.  Algebraic generators (as a ring) of the ring of invariants $k[T]^W$ then are given by generators of the dominant Weyl chamber (and acting on them by $W$, but this doesn't change the relations).  The generators are given by $(1, 1)$ (a root), $(2,0)$ and $(0, 2)$ where $(a, b)$ is $a \omega_1 + b \omega_2$ for $\omega_1, \omega_2$ fundamental weights.  Thus, $k[x^2, xy, y^2]$.
And to end with a question: is there a way to verify this "on geometric points," meaning I want to take $[\lambda_1:\lambda_2:\lambda_3]$ and think of the $S_3$ orbits somehow.  Something like: the (reduced) Hilbert scheme of three points in $\mathbb{C}$ is $\mathbb{C}^3$ modulo a $\mathbb{C}^*$ action with weights $(1,2,3)$?
Also, is the variety $T//W$ known in general for adjoint type algebraic groups?
 A: Your computation is correct except for typos. The ring you are talking about is $k[x^3, xy, y^3]$ (in one place above, you write $(x^2, xy, y^3)$ and in another you write $(x^2, xy, y^2)$); this ring can also be described as $k[A,B,C]/(AC-B^3)$ (not, as you write, $AC-B^2$.) 
The way I would think about this is that $PGL_3 = PSL_3 = SL_3/Z$ where $Z$ is the central subgroup of $SL_3$ generated by $\mathrm{diag}(\omega, \omega, \omega)$, for $\omega$ a primitive third root of unity. Let $T_{simp}$ be the torus in $SL_3$ and $T_{ad}$ be the torus in $PSL_3$; then $T_{ad} = T_{simp}/Z$. Since $Z$ is central in $SL_3$, the $Z$-action on $T_{simp}$ descends to a $Z$-action on $T_{simp}/W$. 
As you say, $T_{simp}/W \cong \mathbb{A}^2$. We can take the coordinates on $\mathbb{A}^2$ to be $e_1$ and $e_2$ where, for $A \in SL_3$, the characteristic polynomial of $A$ is $t^3 - e_1 t^2 + e_2 t -1$. A direct computation shows that $\mathrm{diag}(\omega, \omega, \omega)$ acts by $(e_1, e_2) \mapsto (\omega e_1, \omega^{-1} e_2)$. The invariants for this action are $e_1^3$, $e_1 e_2$ and $e_2^3$, as you describe. You can think of this geometrically as describing $Z \cong \mathbb{Z}/3 \mathbb{Z}$ orbits in $\mathbb{C}^2$; since we are talking about an action of a finite group on an affine variety, closed points of the ring of invariants really do correspond to orbits. 
I don't know a place where $T/W$ is worked out for general semi-simple groups, but the same methods will work. Let $G_{simp}$ be a simply connected group and $G_{ad}$ its adjoint form. Then $G_{simp} = G_{ad}/Z$ for some finite central subgroup $Z$. Let $T_{simp}$ and $T_{ad}$ be the corresponding torii; let $M_{simp}$ and $M_{ad}$ be the character lattices and $N_{simp}$ and $N_{ad}$ the lattices of one parameter subgroups. Then we have a natural isomorphism $N_{ad}/N_{simp} \cong Z$ and hence $M_{simp}/M_{ad}
\cong Z^{\vee}$, where $Z^{\vee}$ is the dual group $\mathrm{Hom}(Z, \mathbb{C}^{\ast})$.
The coordinate ring of $T_{simp}/W$ is, by definition, $k[M_{simp}]^W$. This turns out to be a polynomial ring in $\dim T$ variables. One natural list of generators is to take $\omega_1$, $\omega_2$, ..., $\omega_r \in M_{simp}$ to be the simple weights and take $\chi_k$ to be the character of the irrep with highest weight $\omega_k$; then $k[M_{simp}]^W$ is the polynomial ring generated by the $\chi_k$.
$Z$ acts on this polynomial ring by $z \cdot \chi_k = \omega_k(z) \chi_k$ where here we have used the map $M_{simp} \to Z^{\vee}$ to think of $\omega_k$ as an element of $Z^{\vee}$ and then paired it with $z \in Z$. 
In type $A_n$ (which is to say, $SL_{n+1}$ and $PSL_{n+1}$), the group $Z$ is $\mathbb{Z}/(n+1)$ and we are looking at $Z$ acting on $\mathbb{C}[e_1, e_2, \ldots, e_n]$ by $z e_k \mapsto \zeta^k e_k$ where $\zeta$ is a primitive $n$-th root of $1$. I doubt there is any description of the invariant ring simpler than saying it is the invariant ring. 
The groups $Z$ for all types are listed in this table in the column "fundamental group" (because $Z$ is the fundamental group of $G_{ad}$). It should be straightforward to work out how they act on the rings $k[M_{simp}]^W$. I am pessimistic about finding simple descriptions for the rings of invariants.
A: I want to complete the ad-hoc argument that I had started in the question statement.
Take the 3-torus $T^3 \subset \mathbb{A}^3$.  If one quotients by the $\mathbb{G}_m$ action and then the $S_3$ action (by permuting coordinates) one has $T_{ad}//W$.  I want to take the quotient in the reverse order.
Take homogeneous coordinates $x,y,z$ on $\mathbb{A}^3$.  The quotient variety $T^3//W$ (where $T^3$ is the 3-torus) has coordinate ring $k[x+y+z, xy+yz+xz, xyz, (xyz)^{-1}]$.  Denote the symmetric polynomials $f_1, f_2, f_3$ and quotient by the $\mathbb{G}_m$-action: one wants the degree 0 part of the ring $k[f_1, f_2, f_3, f_3^{-1}]$.  This is exactly the ring
$$k[\frac{f_1^3}{f_3}, \frac{f_2^3}{f_3^2}, \frac{f_1 f_2}{f_3}]$$
and these generators satisfy the relation $AB = C^3$.
