Let $$S = \{z \in \mathbb{C} : |z - 3| \leq 1$$ and $$z(4 + 3i) + \bar{z}(4 - 3i) \leq 24\}$$. If $$\alpha + i\beta$$ is the point in $$S$$ which is closest to $$4i$$, then $$25(\alpha + \beta)$$ is equal to ______.

We are given $$S = \{z \in \mathbb{C} : |z - 3| \leq 1 \text{ and } z(4 + 3i) + \bar{z}(4 - 3i) \leq 24\}$$.

Let $$z = x + iy$$. Then $$\bar{z} = x - iy$$.

The first condition $$|z - 3| \leq 1$$ describes a disk centered at $$(3, 0)$$ with radius 1.

For the second condition:

$$z(4+3i) + \bar{z}(4-3i) = (x+iy)(4+3i) + (x-iy)(4-3i)$$

$$= (4x - 3y + i(3x+4y)) + (4x - 3y - i(3x+4y))$$

$$= 2(4x - 3y) = 8x - 6y$$

So the second condition is: $$8x - 6y \leq 24$$, or $$4x - 3y \leq 12$$.

We need to find the point in $$S$$ closest to $$4i = (0, 4)$$.

The distance from $$(0, 4)$$ to the center $$(3, 0)$$ is $$\sqrt{9 + 16} = 5$$.

The closest point on the circle $$|z-3| = 1$$ to $$(0,4)$$ lies along the line from $$(3,0)$$ to $$(0,4)$$.

The unit vector from $$(3,0)$$ toward $$(0,4)$$ is $$\frac{(-3, 4)}{5}$$.

The closest point on the circle is: $$(3, 0) + 1 \cdot \frac{(-3, 4)}{5} = \left(3 - \frac{3}{5}, \frac{4}{5}\right) = \left(\frac{12}{5}, \frac{4}{5}\right)$$

Checking if this point satisfies $$4x - 3y \leq 12$$:

$$4 \cdot \frac{12}{5} - 3 \cdot \frac{4}{5} = \frac{48 - 12}{5} = \frac{36}{5} = 7.2 \leq 12$$ $$\checkmark$$

So the closest point is $$\alpha + i\beta = \frac{12}{5} + \frac{4}{5}i$$, giving $$\alpha = \frac{12}{5}$$ and $$\beta = \frac{4}{5}$$.

$$25(\alpha + \beta) = 25\left(\frac{12}{5} + \frac{4}{5}\right) = 25 \cdot \frac{16}{5} = 80$$

The correct answer is $$80$$.