Let vec{a}=2 hat{i}+alpha hat{j}+hat{k},vec{b | vedclass

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(D) The area of a parallelogram with diagonals $\vec{d}_1$ and $\vec{d}_2$ is given by $	ext{Area} = \frac{1}{2} |\vec{d}_1 	imes \vec{d}_2|$.Given

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$19$

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(D) The area of a parallelogram with diagonals $\vec{d}_1$ and $\vec{d}_2$ is given by $	ext{Area} = \frac{1}{2} |\vec{d}_1 	imes \vec{d}_2|$.Given diagonals are $\vec{d}_1 = \vec{a}+\vec{b} = (2-1)\hat{i} + \alpha\hat{j} + (1+1)\hat{k} = \hat{i} + \alpha\hat{j} + 2\hat{k}$ and $\vec{d}_2 = \vec{b}+\vec{c} = -\hat{i} + \beta\hat{j} + (1-1)\hat{k} = -\hat{i} + \beta\hat{j}$.Calculating the cross product $\vec{d}_1 	imes \vec{d}_2$:$\vec{d}_1 	imes \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & \alpha & 2 \ -1 & \beta & 0 \end{vmatrix} = \hat{i}(0 - 2\beta) - \hat{j}(0 - (-2)) + \hat{k}(\beta - (-\alpha)) = -2\beta\hat{i} - 2\hat{j} + (\alpha+\beta)\hat{k}$.The area is $\frac{1}{2} \sqrt{(-2\beta)^2 + (-2)^2 + (\alpha+\beta)^2} = \frac{\sqrt{21}}{2}$.Squaring both sides: $4\beta^2 + 4 + \alpha^2 + \beta^2 + 2\alpha\beta = 21$.Given $\alpha\beta = -6$,substitute this into the equation: $\alpha^2 + 5\beta^2 + 2(-6) + 4 = 21 \implies \alpha^2 + 5\beta^2 - 8 = 21 \implies \alpha^2 + 5\beta^2 = 29$.Since $\alpha, \beta$ are integers and $\alpha\beta = -6$,we test pairs: If $\beta=2, \alpha=-3$,then $(-3)^2 + 5(2)^2 = 9 + 20 = 29$ (Valid). If $\beta=-2, \alpha=3$,then $(3)^2 + 5(-2)^2 = 9 + 20 = 29$ (Valid).Let $(\alpha_1, \beta_1) = (-3, 2)$ and $(\alpha_2, \beta_2) = (3, -2)$.Then $\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2 = (-3)^2 + (2)^2 - (3)(-2) = 9 + 4 + 6 = 19$.

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