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(D) Given $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} - \hat{k}$ and $\overrightarrow{b} = -2 \hat{i} + \alpha \hat{j} + \hat{k}$.The area of the

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$14$

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(D) Given $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} - \hat{k}$ and $\overrightarrow{b} = -2 \hat{i} + \alpha \hat{j} + \hat{k}$.The area of the parallelogram is given by $|\vec{a} 	imes \vec{b}|$.Calculating the cross product: $\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \alpha & 2 & -1 \ -2 & \alpha & 1 \end{vmatrix} = \hat{i}(2 + \alpha) - \hat{j}(\alpha - 2) + \hat{k}(\alpha^{2} + 4)$.$|\vec{a} 	imes \vec{b}| = \sqrt{(\alpha + 2)^{2} + (\alpha - 2)^{2} + (\alpha^{2} + 4)^{2}} = \sqrt{\alpha^{2} + 4\alpha + 4 + \alpha^{2} - 4\alpha + 4 + (\alpha^{2} + 4)^{2}} = \sqrt{2(\alpha^{2} + 4) + (\alpha^{2} + 4)^{2}}$.Given $|\vec{a} 	imes \vec{b}| = \sqrt{15(\alpha^{2} + 4)}$,we have $2(\alpha^{2} + 4) + (\alpha^{2} + 4)^{2} = 15(\alpha^{2} + 4)$.Dividing by $(\alpha^{2} + 4)$,we get $2 + (\alpha^{2} + 4) = 15$,so $\alpha^{2} + 4 = 13$,which means $\alpha^{2} = 9$.Now,calculate $|\vec{a}|^{2} = \alpha^{2} + 2^{2} + (-1)^{2} = \alpha^{2} + 5 = 9 + 5 = 14$.$|\vec{b}|^{2} = (-2)^{2} + \alpha^{2} + 1^{2} = 4 + \alpha^{2} + 1 = \alpha^{2} + 5 = 14$.$\vec{a} \cdot \vec{b} = \alpha(-2) + 2(\alpha) + (-1)(1) = -2\alpha + 2\alpha - 1 = -1$.Finally,$2|\vec{a}|^{2} + (\vec{a} \cdot \vec{b})|\vec{b}|^{2} = 2(14) + (-1)(14) = 28 - 14 = 14$.

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