Two adjacent sides of a parallelogram $$PQRS$$ are given by $$\vec{PQ} = \hat{j} + \hat{k}$$ and $$\vec{PS} = \hat{i} - \hat{j}$$. If the side $$PS$$ is rotated about the point $$P$$ by an acute angle $$\alpha$$ in the plane of the parallelogram so that it becomes perpendicular to the side $$PQ$$, then $$\sin^2\left(\frac{5\alpha}{2}\right) - \sin^2\left(\frac{\alpha}{2}\right)$$ is equal to :

The given adjacent sides of parallelogram $$PQRS$$ are
$$\vec{PQ}= \hat{j}+\hat{k}, \qquad \vec{PS}= \hat{i}-\hat{j}.$$

First, find the angle $$\theta$$ between $$\vec{PQ}$$ and $$\vec{PS}$$.
Using the dot-product formula $$\vec{a}\cdot\vec{b}=|\vec{a}|\,|\vec{b}|\cos\theta$$,

$$\vec{PQ}\cdot\vec{PS}= (0,1,1)\cdot(1,-1,0) = 0\cdot1 + 1\cdot(-1) + 1\cdot0 = -1.$$

Magnitudes:
$$|\vec{PQ}| = \sqrt{0^2+1^2+1^2}= \sqrt2, \qquad |\vec{PS}| = \sqrt{1^2+(-1)^2+0^2}= \sqrt2.$$

Therefore,
$$\cos\theta = \frac{-1}{|\vec{PQ}|\,|\vec{PS}|}= \frac{-1}{(\sqrt2)(\sqrt2)}= -\frac12.$$

So $$\theta = 120^{\circ}\;(=\frac{2\pi}{3}).$$

The side $$PS$$ is rotated in the plane of the parallelogram through an acute angle $$\alpha$$ to a new position (call it $$PT$$) such that $$PT \perp PQ$$. Hence the new angle between $$PQ$$ and $$PT$$ must be $$90^{\circ}$$.

Since the original angle is $$120^{\circ}$$ and we need $$90^{\circ}$$, the amount of rotation is
$$\alpha = 120^{\circ}-90^{\circ}=30^{\circ}\;(=\frac{\pi}{6}),$$
which is indeed acute.

Now evaluate $$\sin^2\!\left(\frac{5\alpha}{2}\right) - \sin^2\!\left(\frac{\alpha}{2}\right).$$

Compute half-angles:
$$\frac{\alpha}{2}= \frac{30^{\circ}}{2}=15^{\circ}, \qquad \frac{5\alpha}{2}= \frac{5\cdot30^{\circ}}{2}=75^{\circ}.$$

Using standard values:
$$\sin15^{\circ}= \sin(45^{\circ}-30^{\circ}) = \frac{\sqrt2}{2}\cdot\frac12 - \frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2} = \frac{\sqrt2-\sqrt6}{4},$$
$$\sin75^{\circ}= \sin(45^{\circ}+30^{\circ}) = \frac{\sqrt2}{2}\cdot\frac12 + \frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2} = \frac{\sqrt2+\sqrt6}{4}.$$

Squares:
$$\sin^2 15^{\circ}= \left(\frac{\sqrt2-\sqrt6}{4}\right)^2 = \frac{8-4\sqrt3}{16}= \frac12 - \frac{\sqrt3}{4},$$
$$\sin^2 75^{\circ}= \left(\frac{\sqrt2+\sqrt6}{4}\right)^2 = \frac{8+4\sqrt3}{16}= \frac12 + \frac{\sqrt3}{4}.$$

Desired difference:
$$\sin^2\!\left(\frac{5\alpha}{2}\right) - \sin^2\!\left(\frac{\alpha}{2}\right) = \left(\frac12 + \frac{\sqrt3}{4}\right) - \left(\frac12 - \frac{\sqrt3}{4}\right) = \frac{\sqrt3}{2}.$$

Hence the required value is $$\displaystyle \frac{\sqrt3}{2}.$$
Option B which is: $$\frac{\sqrt3}{2}$$.