Let `P = {z in : |z + 2 – 3i | le 1}` and `Q = {z in C : z (1 + i) + overline z (1 – i) le –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 ________.

Answer

To solve the problem, we need to analyze the sets \( P \) and \( Q \) and find the intersection \( P \cap Q \). We will then determine the points \( z_1 \) and \( z_2 \) where \( |z - (3 - 2i)| \) is maximized and minimized, respectively. Finally, we will compute \( |z_1|^2 + 2|z_2|^2 \) and express it in the form \( \alpha + \beta \sqrt{2} \).

### Step 1: Define the set \( P \)

The set \( P \) is defined as:
\[ P = \{ z \in \mathbb{C} : |z + 2 - 3i| \leq 1 \} \]
This represents a circle in the complex plane with center at \( (-2, 3) \) and radius \( 1 \).

### Step 2: Define the set \( Q \)

The set \( Q \) is defined as:
\[ Q = \{ z \in \mathbb{C} : z(1 + i) + \overline{z}(1 - i) \leq -8 \} \]
Let \( z = x + iy \). Then, \( \overline{z} = x - iy \). Substituting these into the inequality gives:
\[ (x + iy)(1 + i) + (x - iy)(1 - i) \leq -8 \]
Expanding this:
\[ (x + iy + ix - y) + (x - iy - ix - y) \leq -8 \]
Combining like terms:
\[ 2x - 2y \leq -8 \implies x - y \leq -4 \]
This represents a line in the complex plane.

### Step 3: Find the intersection \( P \cap Q \)

We need to find the points where the circle defined by \( P \) intersects with the line defined by \( Q \). The equation of the circle is:
\[ (x + 2)^2 + (y - 3)^2 \leq 1 \]
And the equation of the line is:
\[ x - y = -4 \implies y = x + 4 \]

### Step 4: Substitute the line equation into the circle equation

Substituting \( y = x + 4 \) into the circle equation:
\[ (x + 2)^2 + ((x + 4) - 3)^2 \leq 1 \]
This simplifies to:
\[ (x + 2)^2 + (x + 1)^2 \leq 1 \]
Expanding both squares:
\[ (x^2 + 4x + 4) + (x^2 + 2x + 1) \leq 1 \]
Combining like terms:
\[ 2x^2 + 6x + 5 \leq 1 \implies 2x^2 + 6x + 4 \leq 0 \]
Dividing by 2:
\[ x^2 + 3x + 2 \leq 0 \]
Factoring:
\[ (x + 1)(x + 2) \leq 0 \]
The roots are \( x = -1 \) and \( x = -2 \). The solution is:
\[ -2 \leq x \leq -1 \]

### Step 5: Find corresponding \( y \) values

Using \( y = x + 4 \):
- For \( x = -2 \), \( y = -2 + 4 = 2 \) → Point \( z_1 = -2 + 2i \)
- For \( x = -1 \), \( y = -1 + 4 = 3 \) → Point \( z_2 = -1 + 3i \)

### Step 6: Calculate \( |z_1|^2 \) and \( |z_2|^2 \)

Calculating \( |z_1|^2 \):
\[ |z_1|^2 = |-2 + 2i|^2 = (-2)^2 + (2)^2 = 4 + 4 = 8 \]
Calculating \( |z_2|^2 \):
\[ |z_2|^2 = |-1 + 3i|^2 = (-1)^2 + (3)^2 = 1 + 9 = 10 \]

### Step 7: Compute \( |z_1|^2 + 2|z_2|^2 \)

Now we compute:
\[ |z_1|^2 + 2|z_2|^2 = 8 + 2 \times 10 = 8 + 20 = 28 \]

### Step 8: Express in the form \( \alpha + \beta \sqrt{2} \)

We need to express \( 28 \) in the form \( \alpha + \beta \sqrt{2} \). Here, \( \alpha = 28 \) and \( \beta = 0 \).

Thus, \( \alpha + \beta = 28 + 0 = 28 \).

### Final Answer
\[ \boxed{28} \]