One. (20 points) Let $k$ be a constant. The equation with respect to $x$
$$
x^{2}-2 x+\frac{3 k^{2}-9 k}{x^{2}-2 x-2 k}=3-2 k
$$

has four distinct real roots. Find the range of values for $k$.
Let $y=x^{2}-2 x-2 k$. Then the original equation can be transformed into
$$
\begin{array}{l}
y+2 k+\frac{3 k^{2}-9 k}{y}=3-2 k \\
\Rightarrow y^{2}+(4 k-3) y+3 k^{2}-9 k=0 \\
\Rightarrow(y+3 k)(y+k-3)=0 \\
\Rightarrow y=-3 k \text { or } y=-k+3 \\
\Rightarrow x^{2}-2 x+k=0
\end{array}
$$
$$
\text { or } x^{2}-2 x-k-3=0 \text {. }
$$

Given that the original equation has four distinct real roots, therefore,
$$
\left\{\begin{array}{l}
\Delta_{1}=(-2)^{2}-4 \times 1 \times k>0, \\
\Delta_{2}=(-2)^{2}-4 \times 1 \times(-k-3)>0 .
\end{array}\right.
$$

Solving this, we get $-4<k<1$.
Since $y \neq 0$, we know $k \neq 0$ or 3.
Also, because the roots of equation (1) and equation (2) are different, $-k \neq k+3$, i.e., $k \neq-\frac{3}{2}$.
Therefore, the range of $k$ is
$-4<k<1$ and $k \neq 0$ and $-\frac{3}{2}$.