Let $$P = \{z \in \mathbb{C} : |z + 2 - 3i| \leq 1\}$$ and $$Q = \{z \in \mathbb{C} : z(1+i) + \bar{z}(1-i) \leq -8\}$$. Let in $$P \cap Q$$, $$|z - 3 + 2i|$$ be maximum and minimum at $$z_1$$ and $$z_2$$ respectively. If $$|z_1|^2 + 2|z_2|^2 = \alpha + \beta\sqrt{2}$$, where $$\alpha, \beta$$ are integers, then $$\alpha + \beta$$ equals:

$$|z - (-2 + 3i)| \le 1$$
This represents a circular disc (including the boundary) with:Center $$C = (-2, 3)$$ Radius $$R = 1$$
The condition is $$z(1 + i) + \bar{z}(1 - i) \le -8$$. Let's substitute $$z = x + iy$$ and $$\bar{z} = x - iy$$:
$$(x + iy)(1 + i) + (x - iy)(1 - i) \le -8$$ $$(x + ix + iy - y) + (x - ix - iy - y) \le -8$$
$$2x - 2y \le -8 \implies x - y \le -4 \implies y \ge x + 4$$
This represents the half-plane on and above the straight line L: x - y + 4 = 0.
$$d = \frac{|(-2) - (3) + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$$

Since the distance $$d = \frac{1}{\sqrt{2}} \approx 0.707$$ is less than the radius R = 1, the line cuts through the circle. The region $$P \cap Q$$ is the segment of the circular disc lying above or on the line y = x + 4.

The expression $$|z - (3 - 2i)|$$ represents the distance from a variable point z in $$P \cap Q$$ to a fixed point $$A(3, -2)$$.
Let's find the position of point $$A(3, -2)$$ relative to the line L:
3 - (-2) + 4 = 9 > 0.
the circle center C(-2, 3) into x - y + 4 = -1 < 0.

Since they have opposite signs, A and C lie on opposite sides of the line. 

The maximum distance from A to any point in the circle occurs along the line passing through the center C and A, terminating at the far end of the circle. Let's check if this line segment passes through $$P \cap Q$$
$$m_{CA} = \frac{-2 - 3}{3 - (-2)} = \frac{-5}{5} = -1$$
y - 3 = -1(x + 2) \implies x + y - 1 = 0

Notice that the slope of $CA$ is $-1$, which is perpendicular to the boundary line L(slope = 1).
The furthest point on the circle from $A$ is $$z_1$$, located along the ray $$\overrightarrow{AC}$$ at a distance of CA + R. Because C is well inside the region Q, this furthest point definitely lies in $P \cap Q$.
Let's write $$z_1$$ using parametric form from center C(-2, 3) along the direction away from A.

The vector from A to C moves in the direction of $(-1, 1)$. The unit direction vector is:
$$\hat{u} = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$$
Since $$z_1$$ is at a distance R = 1 from C in this direction:
$$z_1 = \left(-2 - \frac{1}{\sqrt{2}}\right) + i\left(3 + \frac{1}{\sqrt{2}}\right)$$

The closest points to A within $$P \cap Q$$ must lie on the boundary line x - y + 4 = 0 because the entire region $P \cap Q$ is on the other side of this line relative to A. Since line CA is perpendicular to L, the absolute closest point to $A$ on the line is the intersection of CA and L

Let's find this intersection point by solving: y = x + 4 ,x + y - 1 = 0
$$x + (x + 4) - 1 = 0 \implies 2x = -3 \implies x = -\frac{3}{2}$$ $$y = -\frac{3}{2} + 4 = \frac{5}{2}$$
This intersection point is $$M\left(-\frac{3}{2}, \frac{5}{2}\right)$$. Let's check if M lies inside or on the circle:

$$\text{Distance } CM = \sqrt{\left(-2 - \left(-\frac{3}{2}\right)\right)^2 + \left(3 - \frac{5}{2}\right)^2} = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \frac{1}{\sqrt{2}} < 1$$
Since M lies inside the circle, it is a valid point in $$P \cap Q$$. Thus, the minimum distance point is exactly this projection point:
$$z_2 = -\frac{3}{2} + \frac{5}{2}i$$

$$z_1 = \left(-2 - \frac{1}{\sqrt{2}}\right) + i\left(3 + \frac{1}{\sqrt{2}}\right)$$ $$|z_1|^2 = \left(-2 - \frac{1}{\sqrt{2}}\right)^2 + \left(3 + \frac{1}{\sqrt{2}}\right)^2$$ $$|z_1|^2 = \left(4 + 2\sqrt{2} + \frac{1}{2}\right) + \left(9 + 3\sqrt{2} + \frac{1}{2}\right)$$ $$|z_1|^2 = 14 + 5\sqrt{2}$$

$$z_2 = -\frac{3}{2} + \frac{5}{2}i$$ $$|z_2|^2 = \left(-\frac{3}{2}\right)^2 + \left(\frac{5}{2}\right)^2 = \frac{9}{4} + \frac{25}{4} = \frac{34}{4} = \frac{17}{2}$$ $$2|z_2|^2 = 2 \times \frac{17}{2} = 17$$
$$|z_1|^2 + 2|z_2|^2 = (14 + 5\sqrt{2}) + 17 = 31 + 5\sqrt{2}$$