II. (25 points) Given the parabola $y=a x^{2}+b x+2$ intersects the $x$-axis at points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)\left(x_{1}<x_{2}\right)$, and $x_{1}, x_{2}$ are the two real roots of the equation $x^{2}-x-6=0$, $C$ is the intersection point of the parabola with the $y$-axis.
(1) Find the equation of the axis of symmetry of the parabola.
(2) Find the equations of the lines $A C$ and $B C$.
(3) If a moving line $y=m(0<m<2)$ intersects segments $A C$ and $B C$ at points $D$ and $E$ respectively, does there exist a point $P$ on the $x$-axis such that $\triangle D E P$ is an isosceles right triangle? If it exists, find the position of point $P$; if not, explain the reason.
(1) From $x^{2}-x-6=0$, we solve to get $x_{1}=-2, x_{2}=3$.
Thus, $A(-2,0), B(3,0)$.
Therefore, the equation of the axis of symmetry of the parabola is $x=\frac{1}{2}\left(x_{1}+x_{2}\right)=\frac{1}{2}$.
(2) By setting $x=0$ in the equation of the parabola, we get $C(0,2)$.

Then $y_{A C}=x+2, y_{B C}=-\frac{2}{3} x+2$.
(3) Assume there exists a point $P$ on the $x$-axis such that $\triangle D E P$ is an isosceles right triangle.
When $D E$ is a leg, construct $D P_{1} \perp x$-axis at point $P_{1}$.
It is easy to see that $D E=D P_{1}=m$.
By $\triangle C D E \backsim \triangle C A B$
$\Rightarrow \frac{m}{5}=\frac{2-m}{2} \Rightarrow m=\frac{10}{7}$
$\Rightarrow D\left(-\frac{4}{7}, \frac{10}{7}\right) \Rightarrow P_{1}\left(-\frac{4}{7}, 0\right)$.
Similarly, $P_{2}\left(\frac{6}{7}, 0\right)$.
When $D E$ is the base, construct $G P_{3}$
$\perp x$-axis at point $P_{3}$ through the midpoint $G$ of $D E$.
Then $D E=2 G P_{3}=2 m$.
By $\triangle C D E \backsim \triangle C A B$
$\Rightarrow \frac{2 m}{5}=\frac{2-m}{2} \Rightarrow m=\frac{10}{9}$
$\Rightarrow D\left(-\frac{8}{9}, \frac{10}{9}\right), E\left(\frac{4}{3}, \frac{10}{9}\right) \Rightarrow P_{3}\left(\frac{2}{9}, 0\right)$.
In summary, there are three points $P$ that satisfy the conditions:
$$
\left(-\frac{4}{7}, 0\right),\left(\frac{6}{7}, 0\right),\left(\frac{2}{9}, 0\right) \text {. }
$$