In this solution $P$ means $P(x,y,z)$, for a polynomial $P$. Let $C=-B-A$ we get that $$f=A\cdot (x-y)+B\cdot (y-z)+(-B-A)\cdot (z-x)=A\cdot(2x-y-z)+B\cdot (y+x-2z)$$Therefore all we need to show is that there exists a polynomial $B$ such that $A=\frac{B\cdot(y+x-2z)-f}{y-2x+z}$ is a polynomial. Consider $f$ as a polynomial in $y$ and do the division of $f$ by $(y+x-2z)$ , we get that $f=g\cdot(y+x-2z)+h$ were $h$ is a polynomial in $(x,z)$. From the condition that $f(w,w,w)=0$ we have $0=f(w,w,w)=g(w,w,w) \cdot (w+w-2w)+h(w,w)$ wich implies that $h(w,w)=0$ for every real $w$. It is very known that $x-z$ divides $h$,so we let $h=s\cdot(x-z)$ where $s$ is a polynomial in $x,z$. Therefore, we have $$A=-g+\frac{B\cdot(y-2z+x)-s\cdot(x-z)}{y-2x+z}$$. Let $B=B_1 \cdot (y-2x-z)+\frac{s}{3}$ , we have $$A=-g+B_1 \cdot(y-2z+x)+\frac{\frac{s}{3}\cdot(y-2z+x)-s\cdot(x-z)}{y-2x+z}=-g+B_1\cdot(y-2z+x)+s$$wich is clearly a polynomial. And clearly there are infinitely many polynomials $B_1$ we can choose , so there is no polynomial $f$ such that can be uniquely determined. $\blacksquare$