12.15 Let $\approx, w, \lambda$ be complex numbers, $|\lambda| \neq 1$, and consider the equation $\bar{z}-\lambda z=w$ in $z$. There are four conclusions:
I. $z=\frac{\bar{\lambda} w+\bar{w}}{1-|\lambda|^2}$ is a solution to this equation;
II. This equation has only one solution;
III. This equation has two solutions;
N. This equation has infinitely many solutions.

Then $\square$
(A) Only I and II are correct.
(B) Only I and III are correct.
(C) Only I and N are correct.
(D) None of (A), (B), (C) are correct.
(China High School Mathematics League, 1985)
[Solution] From $\bar{z}-\lambda z=w$, we can get
$\overline{\bar{z}-\lambda z}=\bar{w}$, i.e., $z-\bar{\lambda} \bar{z}=\bar{w}$,
and $\bar{\lambda}(\bar{z}-\lambda z)=\bar{\lambda} w$, i.e., $\bar{\lambda} \cdot \bar{z}-|\lambda|^{2} \cdot z=\bar{\lambda} w$,
(1) + (2) gives $\left(1-|\lambda|^{2}\right) \quad z=\bar{\lambda} \cdot w+\bar{w}$.
Given $|\lambda| \neq 1$, i.e., $1-|\lambda|^{2} \neq 0$.
Therefore, $z=\frac{\bar{\lambda} w+\bar{w}}{1-|\lambda|^{2}}$ is the unique solution of the equation.
Hence, the answer is $(A)$.