The equation of the line passing through (3, | vedclass

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(C) Let the direction vectors of the two given lines be $\vec{b_1} = 2\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b_2} = \hat{i} - 2\hat{j} + 2\hat{k}$.Si

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$\frac{x-3}{2} = \frac{y+1}{3} = \frac{z-2}{2}$

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(C) Let the direction vectors of the two given lines be $\vec{b_1} = 2\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b_2} = \hat{i} - 2\hat{j} + 2\hat{k}$.Since the required line is perpendicular to both lines,its direction vector $\vec{b}$ is given by the cross product $\vec{b_1} 	imes \vec{b_2}$.$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 1 \ 1 & -2 & 2 \end{vmatrix} = \hat{i}(-4 + 2) - \hat{j}(4 - 1) + \hat{k}(-4 + 2) = -2\hat{i} - 3\hat{j} - 2\hat{k}$.We can take the direction ratios as $(2, 3, 2)$ by multiplying by $-1$.The line passes through the point $(3, -1, 2)$.Thus,the equation of the line is $\frac{x - 3}{2} = \frac{y - (-1)}{3} = \frac{z - 2}{2}$,which simplifies to $\frac{x - 3}{2} = \frac{y + 1}{3} = \frac{z - 2}{2}$.

The equation of the line passing through $(3, -1, 2)$ and perpendicular to the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2 \hat{i} - 2 \hat{j} + \hat{k})$ and $\vec{r} = (2 \hat{i} + \hat{j} - 3 \hat{k}) + \mu(\hat{i} - 2 \hat{j} + 2 \hat{k})$ is

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