Let $$\vec{a} = \alpha \hat{i} + 2\hat{j} - \hat{k}$$ and $$\vec{b} = -2\hat{i} + \alpha \hat{j} + \hat{k}$$, where $$\alpha \in \mathbf{R}$$. If the area of the parallelogram whose adjacent sides are represented by the vectors $$\vec{a}$$ and $$\vec{b}$$ is $$\sqrt{15(\alpha^2 + 4)}$$, then the value of $$2|\vec{a}|^2 + (\vec{a} \cdot \vec{b})|\vec{b}|^2$$ is equal to

We have $$\vec{a} = \alpha\hat{i} + 2\hat{j} - \hat{k}$$ and $$\vec{b} = -2\hat{i} + \alpha\hat{j} + \hat{k}$$.

Compute $$\vec{a} \times \vec{b}$$:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -1 \\ -2 & \alpha & 1 \end{vmatrix}$$

$$= \hat{i}(2 + \alpha) - \hat{j}(\alpha - 2) + \hat{k}(\alpha^2 + 4)$$

Find the area condition:

$$|\vec{a} \times \vec{b}|^2 = (\alpha + 2)^2 + (\alpha - 2)^2 + (\alpha^2 + 4)^2$$

$$= \alpha^2 + 4\alpha + 4 + \alpha^2 - 4\alpha + 4 + \alpha^4 + 8\alpha^2 + 16$$

$$= \alpha^4 + 10\alpha^2 + 24$$

Given area = $$\sqrt{15(\alpha^2 + 4)}$$, so:

$$|\vec{a} \times \vec{b}|^2 = 15(\alpha^2 + 4) = 15\alpha^2 + 60$$

Setting equal: $$\alpha^4 + 10\alpha^2 + 24 = 15\alpha^2 + 60$$

$$\alpha^4 - 5\alpha^2 - 36 = 0$$

Solve for $$\alpha$$:

Let $$u = \alpha^2$$: $$u^2 - 5u - 36 = 0$$

$$(u - 9)(u + 4) = 0$$

Since $$u = \alpha^2 \geq 0$$: $$\alpha^2 = 9$$, so $$\alpha = \pm 3$$.

Compute the required expression:

$$|\vec{a}|^2 = \alpha^2 + 4 + 1 = 9 + 5 = 14$$

$$|\vec{b}|^2 = 4 + \alpha^2 + 1 = 4 + 9 + 1 = 14$$

$$\vec{a} \cdot \vec{b} = -2\alpha + 2\alpha - 1 = -1$$

$$2|\vec{a}|^2 + (\vec{a} \cdot \vec{b})|\vec{b}|^2 = 2(14) + (-1)(14) = 28 - 14 = 14$$

The correct answer is Option D: $$\boxed{14}$$.