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(B) The given lines are $\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \h

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$\vec{r} \cdot (\hat{i} - \hat{j} - \hat{k}) = 0$

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(B) The given lines are $\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - 2\hat{k})$.These lines pass through the point $\vec{a} = \hat{i} + \hat{j}$ and are parallel to vectors $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{c} = -\hat{i} + \hat{j} - 2\hat{k}$.The normal vector $\vec{n}$ to the plane is given by $\vec{n} = \vec{b} 	imes \vec{c}$.$\vec{n} = (\hat{i} + 2\hat{j} - \hat{k}) 	imes (-\hat{i} + \hat{j} - 2\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -1 \ -1 & 1 & -2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(-2 - 1) + \hat{k}(1 + 2) = -3\hat{i} + 3\hat{j} + 3\hat{k}$.We can take the normal vector as $\vec{n}' = \hat{i} - \hat{j} - \hat{k}$ (dividing by $-3$).The equation of the plane is $(\vec{r} - \vec{a}) \cdot \vec{n}' = 0$,which implies $\vec{r} \cdot \vec{n}' = \vec{a} \cdot \vec{n}'$.$\vec{a} \cdot \vec{n}' = (\hat{i} + \hat{j}) \cdot (\hat{i} - \hat{j} - \hat{k}) = 1 - 1 + 0 = 0$.Thus,the equation of the plane is $\vec{r} \cdot (\hat{i} - \hat{j} - \hat{k}) = 0$.

Find the equation of the plane containing the lines $\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - 2\hat{k})$.

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