Functional equation with three variables

Question

I have a functional equation with three variables. $f(x,y,z)$ is a real function with three variables where y is different from z i.e., $f(x,y,z)$ defined only for $y \neq z$. This function satisfies

1. $f(x,x,y)=0$
2. $f(x,y,x)=1$
3. $f(x,y,z)f(z,y,r)=f(x,y,r)$

What is the general solution for $f$?

Answer 1

There is no solution.

Note that by 3, For all $x,y,z$ you have $$f(x,y,z)=f(x,y,y)f(y,y,z)=f(x,y,y) \cdot 0 =0$$

But this contradicts 2.

Edited If $f(x,y,z)$ is only defined for $y \neq z$ then,a solution is given by $$f(x,y,z)=\frac{x-y}{z-y}$$

Answer 2

Or simpler:

Given $f(x,x,y) = 0$ and $f(x,y,x) = 1$, just set $x=y$ to see the contradiction.

Answer 3

There is no such function. If you put $z=y$ in 3) you get $f(x,y,r)=0$ for all $x,y,r$ (provided $y \neq r$). But this contradicts 2).