11. A. Given the equation in terms of $x$
$$
\frac{x^{2}+k x+3}{x-1}=3 x+k
$$

has only one positive real solution. Find the range of real values for $k$.
Three, 11. A. The original equation can be transformed into
$$
2 x^{2}-3 x-(k+3)=0 \text {. }
$$

When $\Delta=9+8(k+3)=0$, $k=-\frac{33}{8}$, at this time, $x_{1}=x_{2}=\frac{3}{4}$ satisfies the condition;
If $x=1$ is a root of equation (1), then $2 \times 1^{2}-3 \times 1-(k+3)=0 \Rightarrow k=-4$. At this time, the other root of equation (1) is $\frac{1}{2}$. Therefore, the original equation has only one root $x=\frac{1}{2}$. When equation (1) has real roots of opposite signs, $x_{1} x_{2}=\frac{-k+3}{2}-3$. At this time, the original equation has only one positive real root. When equation (1) has one root as 0, $k=-3$, the other root is $x=\frac{3}{2}$, at this time, the original equation has only one positive real root. In summary, the range of $k$ that satisfies the condition is $k=-\frac{33}{8}$ or $k=-4$ or $k \geqslant-3$.