$$
\begin{aligned}
A= & \{z \in \mathbf{C} \mid(a+b \mathrm{i}) \bar{z}+(a-b \mathrm{i}) z+2 c \\
& =0, a, b, c \in \mathbf{R}, ab \neq 0\}, \\
B= & \{z \in \mathbf{C} \mid |z| = 1\} .
\end{aligned}
$$

If $A \cap B=\varnothing$, then with $|a|, |b|, |c|$ as the three sides ( ).
(A) must form an acute triangle
(B) must form a right triangle
(C) must form an obtuse triangle
(D) may not form a triangle
2.D.

Let $z=x+y \mathrm{i} \in A$, then
$$
(a+b \mathrm{i}) \tilde{z}+(a-b \mathrm{i}) z+2 c=0,
$$

we get $(a+b \mathrm{i})(x-y \mathrm{i})+(a-b \mathrm{i})(x+y \mathrm{i})+2 c=0$. 
Expanding and simplifying, we obtain $a x+b y+c=0$.
If $z=x+y \mathrm{i} \in B$, by $|z|=1$, we get $x^{2}+y^{2}=1$.
Thus, $A \cap B=\varnothing$ is equivalent to the line $a x+b y+c=0$ being separated from the circle $x^{2}+y^{2}=1$ in the Cartesian coordinate system. Therefore, the distance from the line $a x+b y+c=0$ to the origin $O(0,0)$ is greater than 1, i.e.,
$$
\frac{|c|}{\sqrt{a^{2}+b^{2}}}>1 \text {. }
$$

Rearranging, we get $a^{2}+b^{2}<c^{2}$. Clearly, $(a, b, c)=(1,1,2)$ satisfies the condition, but in this case, $|a|, |b|, |c|$ cannot form a triangle.
Thus, $|a|, |b|, |c|$ may not form a triangle.