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(A) Let the normal vector to the required plane be $\vec{n} = l\hat{i} + m\hat{j} + n\hat{k}$.Since the plane is perpendicular to $x + 2y + 2z = 5$ an

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$2x - 4y + 3z - 8 = 0$

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(A) Let the normal vector to the required plane be $\vec{n} = l\hat{i} + m\hat{j} + n\hat{k}$.Since the plane is perpendicular to $x + 2y + 2z = 5$ and $3x + 3y + 2z = 8$,the normal vector $\vec{n}$ is perpendicular to the normal vectors $\vec{n}_1 = (1, 2, 2)$ and $\vec{n}_2 = (3, 3, 2)$.Thus,$\vec{n} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 2 \ 3 & 3 & 2 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(2 - 6) + \hat{k}(3 - 6) = -2\hat{i} + 4\hat{j} - 3\hat{k}$.The equation of the plane passing through $(1, -3, -2)$ with normal vector $(-2, 4, -3)$ is:$-2(x - 1) + 4(y + 3) - 3(z + 2) = 0$$-2x + 2 + 4y + 12 - 3z - 6 = 0$$-2x + 4y - 3z + 8 = 0$Multiplying by $-1$,we get $2x - 4y + 3z - 8 = 0$.

The equation of the plane passing through the point $(1, -3, -2)$ and perpendicular to the planes $x + 2y + 2z = 5$ and $3x + 3y + 2z = 8$ is:

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