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(B) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is given by $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.Substituting the point

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$7x - 8y + 3z + 25 = 0$

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(B) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is given by $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.Substituting the point $(-1, 3, 2)$,we get $A(x + 1) + B(y - 3) + C(z - 2) = 0$ ..... $(i)$.Since the plane $(i)$ is perpendicular to the planes $x + 2y + 3z = 5$ and $3x + 3y + z = 0$,its normal vector $\vec{n} = (A, B, C)$ must be perpendicular to the normal vectors $\vec{n_1} = (1, 2, 3)$ and $\vec{n_2} = (3, 3, 1)$.Thus,the normal vector is $\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 3 & 3 & 1 \end{vmatrix} = \hat{i}(2 - 9) - \hat{j}(1 - 9) + \hat{k}(3 - 6) = -7\hat{i} + 8\hat{j} - 3\hat{k}$.Comparing this with $(A, B, C)$,we have $A = -7, B = 8, C = -3$.Substituting these values into equation $(i)$,we get $-7(x + 1) + 8(y - 3) - 3(z - 2) = 0$.$-7x - 7 + 8y - 24 - 3z + 6 = 0$.$-7x + 8y - 3z - 25 = 0$,which is equivalent to $7x - 8y + 3z + 25 = 0$.

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

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