6. Determine $x, y, z$ if

$$
\frac{a y+b x}{x y}=\frac{b z+c y}{y z}=\frac{c x+a z}{z x}=\frac{4 a^{2}+4 b^{2}+4 c^{2}}{x^{2}+y^{2}+z^{2}}, \quad a, b, c \in \mathbb{R}
$$
Solution. The given expression is equivalent to

$$
\frac{a}{x}+\frac{b}{y}=\frac{c}{z}+\frac{b}{y}=\frac{c}{z}+\frac{a}{x}=\frac{4 a^{2}+4 b^{2}+4 c^{2}}{x^{2}+y^{2}+z^{2}}, \quad a, b, c \in \mathbb{R}
$$

From $\frac{a}{x}+\frac{b}{y}=\frac{c}{z}+\frac{b}{y}=\frac{c}{z}+\frac{a}{x}$, we obtain $\frac{a}{x}=\frac{b}{y}=\frac{c}{z}$, from which it follows that $x=\frac{a z}{c}, y=\frac{b z}{c}$. Now,

$$
\frac{a}{x}+\frac{b}{y}=\frac{a}{\frac{a z}{c}}+\frac{b}{\frac{b z}{c}}=\frac{2 c}{z}
$$

On the other hand,

$$
\frac{a}{x}+\frac{b}{y}=\frac{4 a^{2}+4 b^{2}+4 c^{2}}{x^{2}+y^{2}+z^{2}}=\frac{4 a^{2}+4 b^{2}+4 c^{2}}{\frac{a^{2} z^{2}}{c^{2}}+\frac{b^{2} z^{2}}{c^{2}}+z^{2}}=\frac{4 c^{2}\left(a^{2}+b^{2}+c^{2}\right)}{z^{2}\left(a^{2}+b^{2}+c^{2}\right)}=\frac{4 c^{2}}{z^{2}}
$$

From (1) and (2) we have $\frac{2 c}{z}=\frac{4 c^{2}}{z^{2}} \Rightarrow \frac{2 c}{z}=1 \Rightarrow z=2 c$. From here, $x=2 a, y=2 b$.