Can we clear out $a,b,c$ in four equations: $ax+by+cz=0, cx+ay+bz=0, a+b+c=1, a^2+b^2+c^2=1$?
In my search of characteristic cone of $u_{xy}+u_{yz}+u_{zx}=0$, I need to clear out $a,b,c$ in the above formula to get an equation depending only on $x,y,z$, which should be a polynomial of $x,y,z$ of homgeneity $2$ (so that it is a cone).
But I find it is not easy...
An attempt: $ax+by+cz=0$ (1); $cx+ay+bz=0$ (2); and adding up to get $bx+cy+az=0$ (3). Oh, it is wrong....
If $a\cdot (1)+b\cdot(2)+c\cdot (3)$, we find by using $ab+bc+ca=0$ to get $x=0$. Similarly $y=z=0$. Oh my god.!
$$ax+by+cz=0\tag 1$$ $$cx+ay+bz=0\tag 2$$ $$a+b+c=1\tag 3$$ $$a^2+b^2+c^2=1\tag 4$$
From (3) $\quad c=1-a-b\quad$ Put it into (1) , (2) and (4) :
$$ax+by+(1-a-b)z=0\tag 5$$ $$(1-a-b)x+ay+bz=0\tag 6$$ $$a^2+b^2+(1-a-b)^2=1\tag 7$$
From (5) : $\quad b=\frac{-ax+az-z}{y-z}\quad$ Put it into (6) and (7).
$$\left(1-a-\frac{-ax+az-z}{y-z}\right)x+ay+\frac{-ax+az-z}{y-z}z=0\tag 8$$ $$a^2+\left(\frac{-ax+az-z}{y-z}\right)^2+\left(1-a-\frac{-ax+az-z}{y-z}\right)^2=1\tag 9$$
from (8) : $\quad a=\frac{z^2-xy}{x^2+y^2+z^2-xy-yz-xz}\quad$ Put it into (9). After simplification : $$2\:\frac{xy+yz+xz}{x^2+y^2+z^2-xy-yz-xz}=0$$