5. In the complex plane, the vertices $A, B, C$ of $\square A B C$ correspond to the complex numbers $3+2 \mathrm{i}, 3 \mathrm{i}, 2-\mathrm{i}$, respectively. The moving point $P$ corresponds to the complex number $z$. If the equation $|z|^{2}+\alpha z+\bar{\alpha} \bar{z}+\beta=0$ represents the circumcircle of $\square A B C$, where $\alpha, \beta$ are known complex numbers, then the value of $\alpha+\beta$ is ( ).
$$
\text { (A) } 4+\mathrm{i}(\mathrm{B}) 4-\mathrm{i} \text { (C) }-4+\mathrm{i} \text { (D) }-4-\mathrm{i}
$$
5.(C).

It is easy to know that $|A B|=|A C|$, and $A B \perp A C$, so $\square A B C$ is an isosceles right triangle. Its circumcircle equation is
$$
|z-(1+\mathrm{i})|=\sqrt{5} \text { , }
$$

which means $[z-(1+\mathrm{i})][\bar{z}-\overline{(1+\mathrm{i})}]=5$,
or $|z|^{2}-\overline{(1+\mathrm{i})} z-(1+\mathrm{i}) \bar{z}-3=0$. Comparing this with $|z|^{2}+\alpha z+\bar{\alpha} \bar{z}+\beta=0$, we get
$$
\alpha=-\overline{(1+\mathrm{i})}=-1+\mathrm{i}, \beta=-3 \text {. }
$$

Therefore, $\alpha+\beta=-4+\mathrm{i}$.