One. (20 points) Let the quadratic function $y=-x^{2}+(m-2) x$ $+3(m+1)$ have its graph intersect the $x$-axis at points $A, B$ ($A$ is to the left of $B$), and the $y$-axis at point $C$. The product of the lengths of segments $AO$ and $OB$ equals 6 ($O$ is the origin). Connect $AC, BC$. Find the value of $\sin C$.
According to the problem, we have $A O \cdot O B=6$.
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$, then $A O=\left|x_{1}\right|, O B=\left|x_{2}\right|$.
Thus, $\left|x_{1}\right| \cdot\left|x_{2}\right|=6$, which means $\left|x_{1} \cdot x_{2}\right|=6$.
We consider the following two cases:
(1) $x_{1} \cdot x_{2}=6$, that is
$$
x_{1} \cdot x_{2}=-3(m+1)=6 \text {, }
$$

This gives $m=-3, y=-x^{2}-5 x-6$.
Setting $y=0$, we get $x_{1}=-3, x_{2}=-2, A(-3,0), B(-2,0),$
$$
\begin{array}{l}
C(0,-6)
\end{array}
$$
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} A B \cdot O C=\frac{1}{2}(-2+3) \cdot|-6|=3, \\
A C=\sqrt{9+36}=3 \sqrt{5}, \\
B C=\sqrt{4+36}=2 \sqrt{10} .
\end{array}
$$

Thus, $\frac{1}{2} \cdot 3 \sqrt{5} \cdot 2 \sqrt{10} \cdot \sin C=3$,
$$
\sin C=\frac{1}{5 \sqrt{2}}=\frac{\sqrt{2}}{10} \text {. }
$$
$$
\begin{array}{l}
\text { (2) } x_{1} \cdot x_{2}=-6, \text { that is } \\
-3(m+1)=-6,
\end{array}
$$

This gives $m=1, y=-x^{2}-x+6$.
Setting $y=0$, we get $x_{1}=-3, x_{2}=2, A(-3,0),$
$$
\begin{array}{c}
B(2,0), C(0,6) . \\
S_{\triangle A B C}=\frac{1}{2}(2+3) \times 6=15, \\
A C=\sqrt{9+36}=3 \sqrt{5}, \\
B C=\sqrt{4+36}=2 \sqrt{10},
\end{array}
$$

Thus, $\frac{1}{2} \cdot 3 \sqrt{5} \cdot 2 \sqrt{10} \cdot \sin C=15$,
$$
\sin C=\frac{\sqrt{2}}{2} .
$$