Let a unit vector $$\hat{u} = x\hat{i} + y\hat{j} + z\hat{k}$$ make angles $$\frac{\pi}{2}, \frac{\pi}{3}$$ and $$\frac{2\pi}{3}$$ with the vectors $$\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}, \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$$ and $$\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$$ respectively. If $$\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$$, then $$|\hat{u} - \vec{v}|^2$$ is equal to

For the angle $$\frac{\pi}{2}$$ with $$\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$$, we have $$ \frac{x + z}{\sqrt{2}} = \cos\frac{\pi}{2} = 0 \implies x + z = 0 \quad \cdots(1) $$.

For the angle $$\frac{\pi}{3}$$ with $$\frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$$, $$ \frac{y + z}{\sqrt{2}} = \cos\frac{\pi}{3} = \frac{1}{2} \implies y + z = \frac{1}{\sqrt{2}} \quad \cdots(2)$$.

For the angle $$\frac{2\pi}{3}$$ with $$\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$$, $$ \frac{x + y}{\sqrt{2}} = \cos\frac{2\pi}{3} = -\frac{1}{2} \implies x + y = -\frac{1}{\sqrt{2}} \quad \cdots(3)$$.

From (1), $$z = -x$$. Substituting into (2) gives $$y - x = \frac{1}{\sqrt{2}} \quad \cdots(4)$$, and (3) states $$x + y = -\frac{1}{\sqrt{2}} \quad \cdots(5)$$. 

Adding (4) and (5) yields $$2y = 0 \implies y = 0$$. Then (5) implies $$x = -\frac{1}{\sqrt{2}}$$ and (1) gives $$z = \frac{1}{\sqrt{2}}$$, so $$\hat{u} = -\frac{1}{\sqrt{2}}\hat{i} + 0\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$$.

To compute $$|\hat{u} - \vec{v}|^2$$ where $$\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$$, note that

$$\hat{u} - \vec{v} = \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\hat{i} + \left(0 - \frac{1}{\sqrt{2}}\right)\hat{j} + \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\hat{k}$$

$$= -\sqrt{2}\,\hat{i} - \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k}$$

$$ |\hat{u} - \vec{v}|^2 = 2 + \frac{1}{2} + 0 = \frac{5}{2} $$. Thus the correct answer is Option (2): $$\frac{5}{2}$$.