One, (20 points) In the range of positive real numbers, there is only one number that is a solution to the equation $\frac{x^{2}+k x+3}{x-1}=3 x+k$ with respect to $x$. Find the range of real values for $k$.

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One, the original equation is transformed into
$$
2 x^{2}-3 x-(k+3)=0 \text {. }
$$
(1) When $\Delta=0$ for equation (1), that is
$$
(-3)^{2}-4 \times 2 \times[-(k+3)]=0 \text {. }
$$

Solving gives $k=-\frac{33}{8}$.
At this time, the two roots of equation (1) are $x_{1}=x_{2}=\frac{3}{4}$.
Upon verification, $x=\frac{3}{4}$ is a root of the original equation.
(2) When $x=1$ is a root of equation (1),
$$
2 \times 1^{2}-3 \times 1-(k+3)=0 \text {. }
$$

Solving gives $k=-4$.
At this time, the roots of equation (1) are $x=1$ and $x=\frac{1}{2}$.
Upon verification, the root of the original equation is $x=\frac{1}{2}$.
(3) When equation (1) has two roots of opposite signs, we have
$$
x_{1} x_{2}=\frac{-(k+3)}{2} < 0$.
At this time, the original equation has only one root in the positive real number range.
(4) When equation (1) has one root as zero, we have
$$
-(k+3)=0 \text {. }
$$

Solving gives $k=-3$.
The other root of equation (1) is $\frac{3}{2}$.
Upon verification, the original equation has only one positive root, which is $\frac{3}{2}$. In summary, the range of $k$ that satisfies the conditions is $k=-\frac{33}{8}$ or -4 or $k \geqslant-3$.