Let the equation of the curve $\Gamma$ be $F(x, y)=0$, simply denoted as $F=0$ (the same below), and the sides $A_{i} A_{i+1}$ of the inscribed hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ on the curve $\Gamma$ have the equations $f_{i}=0(i=1,2, \cdots, 6)$, and the diagonal $A_{1} A_{4}$ has the equation $g=0$. Then the equation of the conic section passing through the points $A_{1}, A_{2}, A_{3}, A_{4}$ is
$$g f_{2}+\lambda f_{1} f_{3}=0.$$

Since $A_{1}, A_{2}, A_{3}, A_{4}$ are on the curve $\Gamma$, there must exist $\lambda_{1}, \mu_{1}$ such that
$$g f_{2}+\lambda_{1} f_{1} f_{3}=\mu_{1} F=0.$$

Similarly, the equation of the conic section passing through the points $A_{1}, A_{4}, A_{5}, A_{6}$ is $g f_{5}+\lambda f_{4} f_{6}=0$, and there exist $\lambda_{2}, \mu_{2}$ such that
$$g f_{5}+\lambda_{2} f_{4} f_{6}=\mu_{2} F=0.$$

Eliminating $g$ from equations (1) and (2) yields
$$\lambda_{1} f_{1} f_{3} f_{5}-\lambda_{2} f_{2} f_{4} f_{6} =\left(\mu_{1} f_{5}-\mu_{2} f_{2}\right) F=0.$$

Let $A_{1} A_{2} \cap A_{4} A_{5}=P, A_{1} A_{6} \cap A_{4} A_{3}=Q$,
$$A_{2} A_{3} \cap A_{5} A_{6}=R.$$

Since the coordinates of point $P$ satisfy $f_{1}=0, f_{4}=0$, point $P$ lies on the curve (3).
Similarly, points $Q$ and $R$ also lie on the curve (3).
Since $P, Q, R$ are not on the curve $\Gamma$, their coordinates satisfy $F \neq 0, \mu_{1} f_{5}-\mu_{2} f_{2}=0$.
Thus, points $P, Q, R$ are collinear on the line $\mu_{1} f_{5}-\mu_{2} f_{2}=0$.