3. Let $\mathbb{R}_{+}$ be the set of positive real numbers. Find all functions $f: \mathbb{R}_{+}^{3} \rightarrow \mathbb{R}_{+}$ such that for all positive real numbers $x, y, z$ and $k$ the following three conditions hold:

(i) $x f(x, y, z)=z f(z, y, x)$;

(ii) $f\left(x, k y, k^{2} z\right)=k f(x, y, z)$;

(iii) $f(1, k, k+1)=k+1$.

(Great Britain)
3. From the properties of the function $f$, we obtain that for all $x, y, z, a, b > 0$,

$$
f\left(a^{2} x, a b y, b^{2} z\right)=b f\left(a^{2} x, a y, z\right)=b \cdot \frac{z}{a^{2} x} f\left(z, a y, a^{2} x\right)=\frac{b z}{a x} f(z, y, x)=\frac{b}{a} f(x, y, z)
$$

We will choose $a$ and $b$ such that the triplet $(a^{2} x, a b y, b^{2} z)$ is of the form $(1, k, k+1)$ for some $k$: we will take $a=\frac{1}{\sqrt{x}}$ and $b$ such that $b^{2} z - a b y = 1$, which solving the quadratic equation gives $b=\frac{y+\sqrt{y^{2}+4 x z}}{2 z \sqrt{x}}$ and $k=\frac{y\left(y+\sqrt{y^{2}+4 x z}\right)}{2 x z}$. Now we easily obtain

$$
f(x, y, z)=\frac{a}{b} f\left(a^{2} x, a b y, b^{2} z\right)=\frac{a}{b} f(1, k, k+1)=\frac{a}{b}(k+1)=\frac{y+\sqrt{y^{2}+4 x z}}{2 x} .
$$

It is directly verified that $f$ satisfies the conditions of the problem.