For the function $\mathrm{Az} f(x, y, z)$, it holds for any real number $t$ that

$$
f(x+t, y+t, z+t)=t+f(x, y, z) ; f(t x, t y, t z)=t f(x, y, z) ; f(x, y, z)=f(y, x, z)=f(x, z, y)
$$

What is $f(2000,2001,2002) ?$
The third property implies that the function $f$ is symmetric, meaning it takes the same value for any permutation of the variables $x, y, z$.

Thus, for any real number $d$,

$$
f(-d, 0, d)=f(d, 0, -d)
$$

On the other hand, applying the second property for $t=-1$,

$$
f(-d, 0, d)=-f(d, 0, -d)
$$

Comparing the two results,

$$
f(-d, 0, d)=0.
$$

Using the first property, we finally get

$$
f(2000, 2001, 2002)=2001+f(-1, 0, 1)=2001.
$$

Remarks. 1. Similarly, it can be shown that if $x, y$, and $z$ are consecutive terms of an arithmetic sequence such that $y=\frac{x+z}{2}$, then $f(x, y, z)=y$.

2. A function with the properties given in the problem exists, for example, $f(x, y, z)=\frac{x+y+z}{3}$. However, it is not true that the given properties are satisfied only by this function. Rácz Béla András (Fazekas M. Főv. Gyak. Gimn.) found infinitely many such functions. It is easy to verify that for any real $a$, the following functions have the given properties:

$$
f(x, y, z)=\frac{x+y+z}{3}, \text{ if } x, y, z \text{ are not equal; } ax+(1-a)z, \text{ if } x=y; ay+(1-a)x, \text{ if } y=z; az+(1-a)y, \text{ if } x=z.
$$