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(D) Given,$m = \sqrt{3} 	imes 	ext{unit vector of } [(\hat{i}+\hat{j}) 	imes (\hat{j}-\hat{k})]$.First,calculate the cross product: $(\hat{i}+\hat{

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$3\sqrt{2}$

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(D) Given,$m = \sqrt{3} 	imes 	ext{unit vector of } [(\hat{i}+\hat{j}) 	imes (\hat{j}-\hat{k})]$.First,calculate the cross product: $(\hat{i}+\hat{j}) 	imes (\hat{j}-\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 0 \ 0 & 1 & -1 \end{vmatrix} = \hat{i}(-1-0) - \hat{j}(-1-0) + \hat{k}(1-0) = -\hat{i} + \hat{j} + \hat{k}$.The magnitude is $\sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3}$.Thus,$m = \sqrt{3} 	imes \frac{-\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} = -\hat{i} + \hat{j} + \hat{k}$.Similarly,$n = 2\sqrt{6} 	imes 	ext{unit vector of } [(2\hat{i}-\hat{j}) 	imes (\hat{j}+2\hat{k})]$.Calculate the cross product: $(2\hat{i}-\hat{j}) 	imes (\hat{j}+2\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 0 \ 0 & 1 & 2 \end{vmatrix} = \hat{i}(-2-0) - \hat{j}(4-0) + \hat{k}(2-0) = -2\hat{i} - 4\hat{j} + 2\hat{k}$.The magnitude is $\sqrt{(-2)^2 + (-4)^2 + 2^2} = \sqrt{4+16+4} = \sqrt{24} = 2\sqrt{6}$.Thus,$n = 2\sqrt{6} 	imes \frac{-2\hat{i}-4\hat{j}+2\hat{k}}{2\sqrt{6}} = -2\hat{i} - 4\hat{j} + 2\hat{k}$.The area of the triangle with sides $m$ and $n$ is $\frac{1}{2} |m 	imes n|$.$m 	imes n = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & 1 \ -2 & -4 & 2 \end{vmatrix} = \hat{i}(2+4) - \hat{j}(-2+2) + \hat{k}(4+2) = 6\hat{i} + 6\hat{k}$.Area $= \frac{1}{2} |6\hat{i} + 6\hat{k}| = \frac{1}{2} \sqrt{6^2 + 6^2} = \frac{1}{2} 	imes 6\sqrt{2} = 3\sqrt{2}$ sq. units.

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