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(D) 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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$10x - y - 4z = 27$

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(D) 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 $(2, 5, -3)$,we get $a(x - 2) + b(y - 5) + c(z + 3) = 0$ ..... $(1)$.Since this plane is perpendicular to the planes $x + 2y + 2z = 1$ and $x - 2y + 3z = 4$,its normal vector $\vec{n} = (a, b, c)$ must be perpendicular to the normal vectors of the given planes,$\vec{n_1} = (1, 2, 2)$ and $\vec{n_2} = (1, -2, 3)$.Thus,the normal vector $\vec{n}$ is given by the cross product $\vec{n_1} 	imes \vec{n_2}$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 2 \ 1 & -2 & 3 \end{vmatrix} = \hat{i}(6 - (-4)) - \hat{j}(3 - 2) + \hat{k}(-2 - 2) = 10\hat{i} - 1\hat{j} - 4\hat{k}$.So,$a = 10, b = -1, c = -4$.Substituting these values into equation $(1)$:$10(x - 2) - 1(y - 5) - 4(z + 3) = 0$$10x - 20 - y + 5 - 4z - 12 = 0$$10x - y - 4z - 27 = 0$$10x - y - 4z = 27$.

The equation of the plane passing through the point $(2, 5, -3)$ and perpendicular to the planes $x + 2y + 2z = 1$ and $x - 2y + 3z = 4$ is:

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