9. (16 points) In the $x O y$ plane, draw two lines $A_{1} B_{1}$ and $A_{2} B_{2}$ through a point $P$ on the $x$-axis. Points $A_{1}$, $B_{1}$, $A_{2}$, and $B_{2}$ are all on the parabola $\Gamma: y^{2}=x$. Let $A_{1} B_{2}$ and $A_{2} B_{1}$ intersect the $x$-axis at points $S$ and $T$, respectively. Compare $\overrightarrow{O S} \cdot \overrightarrow{O T}$ with $|O P|^{2}$.
II. $9 . \overrightarrow{O S} \cdot \overrightarrow{O T}=|O P|^{2}$.
This is because for $i, j \in\{1,2\}$, we have
$$
\frac{x_{A_{i}}-x_{B_{j}}}{y_{A_{i}}-y_{B_{j}}}=\frac{\left(y_{A_{i}}\right)^{2}-\left(y_{B_{j}}\right)^{2}}{y_{A_{i}}-y_{B_{j}}}=y_{A_{i}}+y_{B_{j}},
$$

which means the equation of the line $A_{i} B_{j}$ is
$$
x=\left(y_{A_{i}}+y_{B_{j}}\right)\left(y-y_{A_{i}}\right)+x_{A_{i}}.
$$

Let $y=0$, we get
$$
x=-\left(y_{A_{i}}+y_{B_{j}}\right) y_{A_{i}}+x_{A_{i}}=-y_{A_{i}} y_{B_{j}},
$$

which means the x-coordinate of the intersection point of the line $A_{i} B_{j}$ with the x-axis is $-y_{A_{i}} y_{B_{j}}$.
$$
\begin{array}{l}
\text { Then } x_{P}=-y_{A_{1}} y_{B_{1}}=-y_{A_{2}} y_{B_{2}}, \\
x_{S}=-y_{A_{1}} y_{B_{2}}, x_{T}=-y_{A_{2}} y_{B_{1}}. \\
\text { Therefore, } \overrightarrow{O S} \cdot \overrightarrow{O T}=\left(-y_{A_{1}} y_{B_{2}}\right)\left(-y_{A_{2}} y_{B_{1}}\right) \\
=\left(-y_{A_{1}} y_{B_{1}}\right)\left(-y_{A_{2}} y_{B_{2}}\right) \\
=\overrightarrow{O P}^{2}=|O P|^{2}.
\end{array}
$$