Let $$\alpha = 8 - 14i$$, $$A = \left\{z \in \mathbb{C} : \frac{\alpha\bar{z} - \bar{\alpha}z}{z^2 - (\bar{z})^2 - 112i} = 1\right\}$$ and $$B = \{z \in \mathbb{C} : |z + 3i| = 4\}$$. Then, $$\sum_{z \in A \cap B} (Re\ z - Im\ z)$$ is equal to ______.

$$z(\alpha - z) - \bar{z}(\bar{\alpha} - \bar{z}) = -112i$$

$$2i\operatorname{Im}\big(z(\alpha - z)\big) = -112i \implies \operatorname{Im}\big(z(\alpha - z)\big) = -56$$

$$z(\alpha - z) = (x + iy)\big((8 - x) - i(14 + y)\big)$$

$$y(8 - x) - x(14 + y) = -56$$

$$8y - xy - 14x - xy = -56 \implies xy + 7x - 4y - 28 = 0$$

$$(x - 4)(y + 7) = 0 \implies x = 4 \text{ or } y = -7 \quad \text{--- (1)}$$

$$x^2 + (y + 3)^2 = 16 \quad \text{--- (2)}$$

Substituting $$x = 4$$ and $$y = 7$$: 

$$\implies \mathbf{(4, -3)}$$

$$ \implies \mathbf{(0, -7)}$$

$$\sum \big(\operatorname{Re}(z) - \operatorname{Im}(z)\big) = [4 - (-3)] + [0 - (-7)] = 7 + 7 = 14$$