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Question

(A) The normal vector to the plane $S_1: x - 2y + 2z + 3 = 0$ is $\vec{n}_1 = \hat{i} - 2\hat{j} + 2\hat{k}$.The normal vector to the plane $S_2: 3x -

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$2x - y - 2z + 3 = 0$

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(A) The normal vector to the plane $S_1: x - 2y + 2z + 3 = 0$ is $\vec{n}_1 = \hat{i} - 2\hat{j} + 2\hat{k}$.The normal vector to the plane $S_2: 3x - 2y + 4z - 4 = 0$ is $\vec{n}_2 = 3\hat{i} - 2\hat{j} + 4\hat{k}$.The normal vector $\vec{n}$ to the required plane is perpendicular to both $\vec{n}_1$ and $\vec{n}_2$,so $\vec{n} = \vec{n}_1 	imes \vec{n}_2$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 2 \ 3 & -2 & 4 \end{vmatrix} = \hat{i}(-8 + 4) - \hat{j}(4 - 6) + \hat{k}(-2 + 6) = -4\hat{i} + 2\hat{j} + 4\hat{k}$.The equation of the plane is $-4(x - 2) + 2(y - 1) + 4(z - 3) = 0$.$-4x + 8 + 2y - 2 + 4z - 12 = 0$.$-4x + 2y + 4z - 6 = 0$.Dividing by $-2$,we get $2x - y - 2z + 3 = 0$.

Find the equation of the plane passing through the point $(2, 1, 3)$ and perpendicular to the planes $x - 2y + 2z + 3 = 0$ and $3x - 2y + 4z - 4 = 0$.

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