On the solution of these equations

Do not forget to see the Good News at the end of the problem.

This problem is linked to the previous one, up to a changes of coordinates. However, that question is actually about only the first three equations here. I add three new equations and hope that we can treat them as a system of 6 variables.

Let $(a,b,c,u,v,w)\in\mathbb{R}^6$,$$T := -a u -b v- c w,\quad S := T + u +v +w.$$

Consider the following 6 equations:$$(1-a)S^2= u^2 -a T^2,$$$$(1-b)S^2 =v^2 -b T^2,$$$$(1-c)S^2=w^2 -c T^2,$$$$\Big( (1-a)^2 S -a^2 T-u\Big)\Big((1-b)^2 S -b^2 T-v\Big)=\Big((1-a)(1-b)S-abT\Big)^2,$$$$\Big( (1-a)^2 S -a^2 T-u\Big)\Big((1-c)^2 S -c^2 T-w\Big)=\Big((1-a)(1-c)S-acT\Big)^2,$$$$\Big( (1-b)^2 S -b^2 T-v\Big)\Big((1-c)^2 S -c^2 T-w\Big)=\Big((1-b)(1-c)S-bcT\Big)^2.$$I want to find the solutions.

Original problem is:Consider the homogeneous function$$f(u,v,w) = (u+v + w +T)^3 - (u^3 + v^3 + w^3 + T^3).$$I want to show if $x_0$ is a critical point $f$, that is, $\nabla f(x_0)=0$, then $\nabla^2 f(x_0)$ has rank 2, where $\nabla^2 f$ is the Hessian of $f$. So I get the six equation above.

How can I figure it out? Maybe the symmetry is helpful.

Good News: After expressing $a,b,c$ in terms of $(T,S,u,v,w)$ from the first three equations, and then put them into the left three equations to get a system of 4 equations on $(T, u, v, w)$. Mathematica tells me that the non-zero solutions of the original 6 equations must have the form$$(1,0,0,u,0,0)\ \ \mbox{or}\ \ (0,1,0,0,v,0)\ \ \mbox{or} \ \ (0,0,1,0,0,w).$$In fact, the four new equations for $(T,u,v,w)$ now are:$$(eq.1)\ \ S(v(T^2-u^2)^2+ u(T^2-v^2)^2)+ (u^2-v^2)^2 ST = uv(T^2-S^2)^2+ T((u^2-S^2)^2 v + (v^2-S^2)^2 u)$$$$(eq.2)\ \ S(w(T^2-u^2)^2+ u(T^2-w^2)^2)+ (u^2-w^2)^2 ST = uw(T^2-S^2)^2+ T((u^2-S^2)^2 w + (w^2-S^2)^2 u)$$$$(eq.3)\ \ S(w(T^2-v^2)^2+ v(T^2-w^2)^2)+ (v^2-w^2)^2 ST = vw(T^2-S^2)^2+ T((v^2-S^2)^2 w + (w^2-S^2)^2 v)$$$$(eq.4) \ \ S^3 = T^3 + u^3 + v^3 + w^3,\ \ \mbox{where} \ S= T+u+v+w.$$You can only work on them, Mathematica tells me the real solution of this system must take the form, e.g. [$T=-u$, $v=w=0$] or [$T=w=0$, $u=-v$] (other solution are similar by symmetry).