9. (ROM 2) Let $O x, O y, O z$ be three rays, and $G$ a point inside the trihedron $O x y z$. Consider all planes passing through $G$ and cutting $O x, O y, O z$ at points $A, B, C$, respectively. How is the plane to be placed in order to yield a tetrahedron $O A B C$ with minimal perimeter?
9. Let $a, b, c$ be vectors going along $O x, O y, O z$, respectively, such that $\overrightarrow{O G}=$ $a+b+c$. Now let $A \in O x, B \in O y, C \in O z$ and let $\overrightarrow{O A}=\alpha a, \overrightarrow{O B}=\beta b$, $\overrightarrow{O C}=\gamma c$, where $\alpha, \beta, \gamma>0$. Point $G$ belongs to a plane $A B C$ with $A \in O x, B \in O y, C \in O z$ if and only if there exist positive real numbers $\lambda, \mu, \nu$ with sum 1 such that $\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}+\nu \overrightarrow{O C}=\overrightarrow{O G}$, which is equivalent to $\lambda \alpha=\mu \beta=\nu \eta=1$. Such $\lambda, \mu, \nu$ exist if and only if
$$
\alpha, \beta, \gamma>0 \text { and } \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=1 .
$$

Since the volume of $O A B C$ is proportional to the product $\alpha \beta \gamma$, it is minimized when $\frac{1}{\alpha} \cdot \frac{1}{3} \cdot \frac{1}{\gamma}$ is maximized, which occurs when $\alpha=\beta=\gamma=3$ and $G$ is the centroid of $\triangle A B C$.