Let L_1: frac{x+1}{3}=frac{y+2}{2}=frac{z+1}{ | vedclass

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(C) The lines $L_1$ and $L_2$ are parallel to the vectors $\vec{b}_1 = 3 \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b}_2 = 2 \hat{i} + \hat{j} + 3 \hat{

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$\frac{5 \hat{i}-7 \hat{j}-\hat{k}}{5 \sqrt{3}}$

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(C) The lines $L_1$ and $L_2$ are parallel to the vectors $\vec{b}_1 = 3 \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b}_2 = 2 \hat{i} + \hat{j} + 3 \hat{k}$ respectively. The unit vector perpendicular to both $L_1$ and $L_2$ is given by $\hat{n} = \frac{\vec{b}_1 	imes \vec{b}_2}{|\vec{b}_1 	imes \vec{b}_2|}$. First,we calculate the cross product $\vec{b}_1 	imes \vec{b}_2$: $\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 1 \ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(9-2) + \hat{k}(3-4) = 5 \hat{i} - 7 \hat{j} - \hat{k}$. Next,we find the magnitude of the cross product: $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{5^2 + (-7)^2 + (-1)^2} = \sqrt{25 + 49 + 1} = \sqrt{75} = 5 \sqrt{3}$. Therefore,the required unit vector is $\hat{n} = \frac{5 \hat{i} - 7 \hat{j} - \hat{k}}{5 \sqrt{3}}$.

Let $L_1: \frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1}$ and $L_2: \frac{x-2}{2}=\frac{y+2}{1}=\frac{z-3}{3}$ be the given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

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