16. Given in the complex plane $\triangle A B C$ has vertices $A, B, C$ corresponding to the complex numbers $3+2 \mathrm{i}, 3 \mathrm{i}, 2-\mathrm{i}$, and the moving point $P$ corresponds to the complex number $z$. If the equation $|\bar{z}|^{2}+\alpha z+\bar{\alpha} \bar{z}+\beta=0$ represents the circumcircle of $\triangle A B C$, find the complex numbers $\alpha, \beta$.
16. It is known that $|A B|=|A C|$ and $|A B|^{2}+|A C|^{2}=|B C|^{2}$, so $\triangle A B C$ is an isosceles right triangle.

Let $z=x+y i(x, y \in \mathbf{R})$, then the equation of the circle with the hypotenuse $B C$ as the diameter is $x(x-2)+(y+1)(y-3)=0$, i.e., $x^{2}+y^{2} \cdots 2 x-2 y-3=0$. Let $\alpha=a+b \mathrm{i}, \beta=c+d \mathrm{i}(a, b, c, d \in \mathbf{R})$, substituting into the known equation, and transforming it, we get $x^{2}+y^{2}+$ $2 a x-2 b y+c+d \mathrm{i}=0$. Therefore, $d=0$ and $x^{2}+y^{2}+2 a x-2 b y+c=0$. Comparing coefficients, we get $a=-1, b=1$, $c=-3$. Hence, $\alpha=-1+\mathrm{i}, \beta=-3$.