XV OBM

Let $S$ be the sphere that contains the circumcircle of the polygon and $Q$ and let $Q^{\prime }$ be the point diametrically opposite to $Q$ in $S$. Then $\angle Q^{\prime }{P}_{i}Q=90^{\circ }$, that is, $Q^{\prime }{P}_i$ is perpendicular to $Q{P}_i$ and since ${\beta }_i$ contains all lines perpendicular to $Q{P}_i$ passing through $P_i$, $Q^{\prime }$ belongs to ${\beta }_i$.

It remains to prove that the intersection of the planes is not a line. Let $t$ be the intersection of ${\beta }_1$ and ${\beta }_2$. It is a line, since the intersection is non-empty. This line is perpendicular to both $Q{P}_1$ and $Q{P}_2$, so it is perpendicular to plane $Q{P}_1{P}_2$. Analogously, the intersection $u$ of ${\beta }_2$ and ${\beta }_3$ is perpendicular to the plane $Q{P}_2{P}_3$. Since the planes $Q{P}_1{P}_2$ and $Q{P}_2{P}_3$ are secant in $Q{P}_2$, $t\ne u$ and the intersection of all planes is the intersection of $t$ and $u$, which is a single point.