Let $$\vec{a} = 2\hat{i} + \alpha\hat{j} + \hat{k}, \vec{b} = -\hat{i} + \hat{k}, \vec{c} = \beta\hat{j} - \hat{k}$$, where $$\alpha$$ and $$\beta$$ are integers and $$\alpha\beta = -6$$. Let the values of the ordered pair $$(\alpha, \beta)$$, for which the area of the parallelogram of diagonals $$\vec{a} + \vec{b}$$ and $$\vec{b} + \vec{c}$$ is $$\frac{\sqrt{21}}{2}$$, be $$(\alpha_1, \beta_1)$$ and $$(\alpha_2, \beta_2)$$. Then $$\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2$$ is equal to

Given: $$\vec{a} = 2\hat{i} + \alpha\hat{j} + \hat{k}$$, $$\vec{b} = -\hat{i} + \hat{k}$$, $$\vec{c} = \beta\hat{j} - \hat{k}$$, with $$\alpha\beta = -6$$.

Diagonals: $$\vec{a} + \vec{b} = (1, \alpha, 2)$$ and $$\vec{b} + \vec{c} = (-1, \beta, 0)$$.

Area of parallelogram with diagonals $$\vec{d_1}$$ and $$\vec{d_2}$$ = $$\frac{1}{2}|\vec{d_1} \times \vec{d_2}|$$.

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 2 \\ -1 & \beta & 0 \end{vmatrix} = \hat{i}(0-2\beta) - \hat{j}(0+2) + \hat{k}(\beta+\alpha)$$

$$= (-2\beta, -2, \alpha+\beta)$$

$$|\vec{d_1} \times \vec{d_2}|^2 = 4\beta^2 + 4 + (\alpha+\beta)^2$$

Area = $$\frac{\sqrt{21}}{2}$$, so $$\frac{1}{4}[4\beta^2 + 4 + (\alpha+\beta)^2] = \frac{21}{4}$$.

$$4\beta^2 + 4 + \alpha^2 + 2\alpha\beta + \beta^2 = 21$$

$$\alpha^2 + 5\beta^2 + 2(-6) + 4 = 21$$ (using $$\alpha\beta = -6$$)

$$\alpha^2 + 5\beta^2 = 29$$

With $$\alpha\beta = -6$$ (integer solutions): possible $$(\alpha, \beta)$$ pairs: $$(1$$, $$-6)$$, $$(-1$$, $$6)$$, $$(2$$, $$-3)$$, $$(-2$$, $$3)$$, $$(3$$, $$-2)$$, $$(-3$$, $$2)$$, $$(6$$, $$-1)$$, $$(-6$$, $$1)$$.

Check $$\alpha^2 + 5\beta^2 = 29$$:

$$(\alpha_1, \beta_1) = (3, -2)$$ and $$(\alpha_2, \beta_2) = (-3, 2)$$.

$$\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2 = 9 + 4 - (-3)(2) = 13 + 6 = 19$$.

The correct answer is Option A: 19