Let $$A = \{z \in \mathbb{C} : |z - 2 - i| = 3\}$$, $$B = \{z \in \mathbb{C} : \text{Re}(z - iz) = 2\}$$ and $$S = A \cap B$$. Then $$\displaystyle\sum_{z \in S} |z|^2$$ is equal to ________.

The complex numbers in set $$A$$ satisfy the circle equation
$$|z-(2+i)|=3 \; \Longrightarrow \; (x-2)^2+(y-1)^2=9$$ where $$z=x+iy$$.

The numbers in set $$B$$ satisfy
$$\text{Re}(z-iz)=2.$$ First write $$z-iz=z(1-i)=(x+iy)(1-i).$$

Expanding,
$$(x+iy)(1-i)=x+iy-ix-i^2y=(x+y)+i(y-x).$$
Hence $$\text{Re}(z-iz)=x+y,$$ so $$B$$ is the straight line
$$x+y=2.$$

To find the intersection $$S=A\cap B,$$ solve the system
$$\begin{cases}(x-2)^2+(y-1)^2=9 \\[4pt] x+y=2\end{cases}$$

From the line, $$y=2-x.$$ Substitute into the circle:

$$(x-2)^2+\bigl((2-x)-1\bigr)^2=9\\ (x-2)^2+(1-x)^2=9.$$

Expand and collect terms:
$$(x^2-4x+4)+(x^2-2x+1)=9\\ 2x^2-6x+5=9\\ 2x^2-6x-4=0\\ x^2-3x-2=0.$$

Solve the quadratic:
$$x=\frac{3\pm\sqrt{9+8}}{2}=\frac{3\pm\sqrt{17}}{2}.$$

Using $$y=2-x$$, we get the two intersection points

$$z_1=\frac{3+\sqrt{17}}{2}+i\,\frac{1-\sqrt{17}}{2},\qquad z_2=\frac{3-\sqrt{17}}{2}+i\,\frac{1+\sqrt{17}}{2}.$$

Now compute $$|z|^2=x^2+y^2$$ for each point.

For $$z_1$$: $$x_1=\frac{3+\sqrt{17}}{2},\;y_1=\frac{1-\sqrt{17}}{2}$$ $$|z_1|^2=\left(\frac{3+\sqrt{17}}{2}\right)^2+\left(\frac{1-\sqrt{17}}{2}\right)^2 =\frac{(3+\sqrt{17})^2+(1-\sqrt{17})^2}{4} =\frac{26+6\sqrt{17}+18-2\sqrt{17}}{4} =11+\sqrt{17}.$$

For $$z_2$$: $$x_2=\frac{3-\sqrt{17}}{2},\;y_2=\frac{1+\sqrt{17}}{2}$$ $$|z_2|^2=\left(\frac{3-\sqrt{17}}{2}\right)^2+\left(\frac{1+\sqrt{17}}{2}\right)^2 =11-\sqrt{17}.$$

Finally,
$$\sum_{z\in S}|z|^2=(11+\sqrt{17})+(11-\sqrt{17})=22.$$

Therefore the required sum is $$22$$.