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(D) The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula: $	ext{Area} = \frac{1}{2} |\vec{d_1} 	imes \vec{

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$-4$

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(D) The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula: $	ext{Area} = \frac{1}{2} |\vec{d_1} 	imes \vec{d_2}|$. Given $\vec{d_1} = 3 \hat{i} + \lambda \hat{j} + 2 \hat{k}$ and $\vec{d_2} = \hat{i} - 2 \hat{j} + 3 \hat{k}$. First,calculate the cross product $\vec{d_1} 	imes \vec{d_2}$: $\vec{d_1} 	imes \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & \lambda & 2 \ 1 & -2 & 3 \end{vmatrix} = \hat{i}(3\lambda + 4) - \hat{j}(9 - 2) + \hat{k}(-6 - \lambda) = (3\lambda + 4) \hat{i} - 7 \hat{j} - (6 + \lambda) \hat{k}$. The magnitude is $|\vec{d_1} 	imes \vec{d_2}| = \sqrt{(3\lambda + 4)^2 + (-7)^2 + (-(6 + \lambda))^2} = \sqrt{9\lambda^2 + 24\lambda + 16 + 49 + 36 + 12\lambda + \lambda^2} = \sqrt{10\lambda^2 + 36\lambda + 101}$. Given $\frac{1}{2} \sqrt{10\lambda^2 + 36\lambda + 101} = \frac{\sqrt{117}}{2}$,so $10\lambda^2 + 36\lambda + 101 = 117$. $10\lambda^2 + 36\lambda - 16 = 0 \implies 5\lambda^2 + 18\lambda - 8 = 0$. Factoring the quadratic: $5\lambda^2 + 20\lambda - 2\lambda - 8 = 0 \implies 5\lambda(\lambda + 4) - 2(\lambda + 4) = 0$. $(5\lambda - 2)(\lambda + 4) = 0$. Thus,$\lambda = -4$ or $\lambda = 0.4$. Comparing with the given options,$\lambda = -4$ is the correct choice.

If the area of a parallelogram whose diagonals are represented by vectors $\vec{d_1} = 3 \hat{i} + \lambda \hat{j} + 2 \hat{k}$ and $\vec{d_2} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ is $\frac{\sqrt{117}}{2}$ sq. units,then $\lambda=$

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