265. Three mutually perpendicular lines pass through the point $O, A, B$ and $C$ - points on these lines such that

$$
|O A|=|O B|=|O C|
$$

Let $l$ be an arbitrary line passing through $O$; $A_{1}, B_{1}$ and $C_{1}$ - points symmetric to $A, B$ and $C$ with respect to $l$. Through $A_{1}, B_{1}$ and $C_{1}$, three planes are drawn perpendicular to the lines $O A, O B$ and $O C$, respectively. Find the geometric locus of the points of intersection of these planes.
265. We will consider the given straight lines as coordinate axes. Let the straight line form angles $\alpha, \beta$ and $\gamma$ with these axes. Then the projections of vectors $\overrightarrow{O A_{i}}, \overrightarrow{O B_{1}}, \overrightarrow{O C} \vec{O}_{i}$ on the axes $O A, O B$ and $O C$ respectively will be equal to $a \cos 2 \alpha, a \cos 2 \beta, a \cos 2 \gamma, a=|O A|$. Therefore, $M$ the point of intersection of planes passing through $A_{i}, B_{1}, C_{\mathrm{i}}$ respectively perpendicular to $O A, O B, O C,$ will have coordinates ( $a \cos 2 \alpha, a \cos 2 \beta, a \cos 2 \gamma$ ). The set of points with coordinates $\left(\cos ^{2} \alpha, \cos ^{2} \beta, \cos ^{2} \gamma\right)$ is a triangle with vertices at the ends of the unit vectors of the axes. Therefore, the sought geometric locus of points is also a triangle, the vertices of which have coordinates ( $-a,-a, a) ;(-a, a,-a) ;(a,-a,-a)$.