If $$\vec{P} = 3\hat{i} + \sqrt{3}\hat{j} + 2\hat{k}$$ and $$\vec{Q} = 4\hat{i} + \sqrt{3}\hat{j} + 2.5\hat{k}$$, then, the unit vector in the direction of $$\vec{P} \times \vec{Q}$$ is $$\frac{1}{x}(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k})$$. The value of $$x$$ is

We are given $$\vec{P} = 3\hat{i} + \sqrt{3}\hat{j} + 2\hat{k}$$ and $$\vec{Q} = 4\hat{i} + \sqrt{3}\hat{j} + 2.5\hat{k}$$.

Compute $$\vec{P} \times \vec{Q}$$.

$$ \vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & \sqrt{3} & 2 \\ 4 & \sqrt{3} & 2.5 \end{vmatrix} $$

$$ = \hat{i}(\sqrt{3} \cdot 2.5 - 2 \cdot \sqrt{3}) - \hat{j}(3 \cdot 2.5 - 2 \cdot 4) + \hat{k}(3\sqrt{3} - 4\sqrt{3}) $$

$$ = \hat{i}(2.5\sqrt{3} - 2\sqrt{3}) - \hat{j}(7.5 - 8) + \hat{k}(-\sqrt{3}) $$

$$ = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j} - \sqrt{3}\hat{k} $$

Find the magnitude.

$$ |\vec{P} \times \vec{Q}| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + (-\sqrt{3})^2} = \sqrt{\frac{3}{4} + \frac{1}{4} + 3} = \sqrt{4} = 2 $$

Find the unit vector.

$$ \hat{n} = \frac{\vec{P} \times \vec{Q}}{|\vec{P} \times \vec{Q}|} = \frac{1}{2}\left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j} - \sqrt{3}\hat{k}\right) = \frac{\sqrt{3}}{4}\hat{i} + \frac{1}{4}\hat{j} - \frac{\sqrt{3}}{2}\hat{k} $$

Compare with the given form.

The unit vector is given as $$\frac{1}{x}(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k})$$.

$$ \frac{1}{x}(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k}) = \frac{\sqrt{3}}{4}\hat{i} + \frac{1}{4}\hat{j} - \frac{\sqrt{3}}{2}\hat{k} $$

Comparing the $$\hat{j}$$ components: $$\frac{1}{x} = \frac{1}{4}$$, so $$x = 4$$.

Verification with $$\hat{i}$$ component: $$\frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4}$$

Verification with $$\hat{k}$$ component: $$\frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}$$

The value of $$x$$ is $$\boxed{4}$$.