Brazilian Mathematical Olympiad

Let $S$ be the sphere that contains the circumcircle of the polygon and $Q$ and let $Q'$ be the point diametrically opposite to $Q$ in $S$. Then $\angle Q'P_iQ = 90^\circ$, that is, $Q'P_i$ is perpendicular to $QP_i$ and since $\beta_i$ contains all lines perpendicular to $QP_i$ passing through $P_i$, $Q'$ 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 $QP_1$ and $QP_2$, so it is perpendicular to plane $QP_1P_2$. Analogously, the intersection $u$ of $\beta_2$ and $\beta_3$ is perpendicular to the plane $QP_2P_3$. Since the planes $QP_1P_2$ and $QP_2P_3$ are secant in $QP_2$, $t \neq u$ and the intersection of all planes is the intersection of $t$ and $u$, which is a single point.