Consider the following two statements :

Statement I: For any two non-zero complex numbers $$z_1, z_2,(|z_1|+|z_2|)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right) \text {, and }$$

Statement II : If $$x, y, z$$ are three distinct complex numbers and $$\mathrm{a}, \mathrm{b}, \mathrm{c}$$ are three positive real numbers such that $$\frac{\mathrm{a}}{|y-z|}=\frac{\mathrm{b}}{|z-x|}=\frac{\mathrm{c}}{|x-y|}$$, then $$\frac{\mathrm{a}^2}{y-z}+\frac{\mathrm{b}^2}{z-x}+\frac{\mathrm{c}^2}{x-y}=1$$.

Between the above two statements,

The area (in sq. units) of the region $$S=\{z \in \mathbb{C}:|z-1| \leq 2 ;(z+\bar{z})+i(z-\bar{z}) \leq 2, \operatorname{lm}(z) \geq 0\}$$ is

Let $$\alpha$$ and $$\beta$$ be the sum and the product of all the non-zero solutions of the equation $$(\bar{z})^2+|z|=0, z \in C$$. Then $$4(\alpha^2+\beta^2)$$ is equal to :