Consider the following two statements : Statement I : For any two non-zero complex numbers $$z_1, z_2$$, $$(|z_1| + |z_2|)\left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right| \leq 2(|z_1| + |z_2|)$$. Statement II : If $$x, y, z$$ are three distinct complex numbers and $$a, b, c$$ are three positive real numbers such that $$\frac{a}{|y-z|} = \frac{b}{|z-x|} = \frac{c}{|x-y|}$$, then $$\frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y} = 1$$. Between the above two statements,

Statement I: $$(|z_1| + |z_2|)\left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right| \leq 2(|z_1| + |z_2|)$$.

Let $$\hat{z_1} = \frac{z_1}{|z_1|}$$ and $$\hat{z_2} = \frac{z_2}{|z_2|}$$. These are unit complex numbers, so $$|\hat{z_1}| = |\hat{z_2}| = 1$$.

$$|\hat{z_1} + \hat{z_2}| \leq |\hat{z_1}| + |\hat{z_2}| = 2$$.

So $$(|z_1| + |z_2|)|\hat{z_1} + \hat{z_2}| \leq 2(|z_1| + |z_2|)$$. Statement I is correct.

Statement II: With $$\frac{a}{|y-z|} = \frac{b}{|z-x|} = \frac{c}{|x-y|} = k$$ (say), so $$a = k|y-z|$$, etc.

$$\frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y} = k^2\left(\frac{|y-z|^2}{y-z} + \frac{|z-x|^2}{z-x} + \frac{|x-y|^2}{x-y}\right)$$

$$= k^2(\overline{y-z} + \overline{z-x} + \overline{x-y})$$ (since $$\frac{|w|^2}{w} = \bar{w}$$)

$$= k^2 \cdot \overline{(y-z)+(z-x)+(x-y)} = k^2 \cdot \overline{0} = 0$$, not 1.

So Statement II claims this equals 1, but it actually equals 0. Statement II is incorrect.

The correct answer is Option (1): Statement I is correct but Statement II is incorrect.