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(A) The normal vectors of the given planes are $\vec{n_1} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{n_2} = 3\hat{i} + 3\hat{j} + 2\hat{k}$.Since the

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$2x - 4y + 3z + 3 = 0$

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(A) The normal vectors of the given planes are $\vec{n_1} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{n_2} = 3\hat{i} + 3\hat{j} + 2\hat{k}$.Since the required plane is perpendicular to both,its normal vector $\vec{n}$ is given by the cross product $\vec{n_1} 	imes \vec{n_2}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 2 \ 3 & 3 & 2 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(2 - 6) + \hat{k}(3 - 6) = -2\hat{i} + 4\hat{j} - 3\hat{k}$.We can take the normal vector as $\vec{n} = 2\hat{i} - 4\hat{j} + 3\hat{k}$.The equation of the plane passing through $(x_0, y_0, z_0) = (1, 2, 1)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.$2(x - 1) - 4(y - 2) + 3(z - 1) = 0$.$2x - 2 - 4y + 8 + 3z - 3 = 0$.$2x - 4y + 3z + 3 = 0$.

The equation of the plane passing through the point $(1, 2, 1)$ and perpendicular to the planes $x + 2y + 2z - 7 = 0$ and $3x + 3y + 2z - 5 = 0$ is:

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