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(C) The equation of a plane passing through the point $(1, 2, -1)$ with normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is $a(x-1) + b(y-2) +

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$r \cdot(2 \hat{i}-7 \hat{j}-13 \hat{k})=1$

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(C) The equation of a plane passing through the point $(1, 2, -1)$ with normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is $a(x-1) + b(y-2) + c(z+1) = 0$.Since the plane is perpendicular to the line of intersection of the planes $r \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $r \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$,the normal vector $\vec{n}$ must be parallel to the cross product of the normals of the two given planes,$\vec{n_1} = 3\hat{i}-\hat{j}+\hat{k}$ and $\vec{n_2} = \hat{i}+4\hat{j}-2\hat{k}$.$\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 1 \ 1 & 4 & -2 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(-6-1) + \hat{k}(12+1) = -2\hat{i} + 7\hat{j} + 13\hat{k}$.Thus,the direction ratios are $(a, b, c) = (-2, 7, 13)$.Substituting these into the plane equation: $-2(x-1) + 7(y-2) + 13(z+1) = 0$.$-2x + 2 + 7y - 14 + 13z + 13 = 0$.$-2x + 7y + 13z + 1 = 0$,which simplifies to $2x - 7y - 13z = 1$.In vector form,this is $r \cdot(2 \hat{i}-7 \hat{j}-13 \hat{k})=1$.

The equation of the plane passing through the point $\hat{i}+2 \hat{j}-\hat{k}$ and perpendicular to the line of intersection of the planes $r \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $r \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is:

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