Equation of Plane Perpendicular to Two Given Planes

To find the equation of the plane passing through the point $(1, 1, 1)$ and perpendicular to the given planes, we first determine the normal vector of the required plane.

Let the equations of the given planes be: $P_1 : 2x + y - 2z = 5$ $P_2 : 3x - 6y - 2z = 7$

The normal vectors to $P_1$ and $P_2$ are: $\vec{n}_1 = 2\hat{i} + \hat{j} - 2\hat{k}$ $\vec{n}_2 = 3\hat{i} - 6\hat{j} - 2\hat{k}$

Since the required plane is perpendicular to both $P_1$ and $P_2$, its normal vector $\vec{n}$ must be perpendicular to both $\vec{n}_1$ and $\vec{n}_2$. Thus, $\vec{n}$ is parallel to the cross product $\vec{n}_1 \times \vec{n}_2$:

$$\vec{n} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 3 & -6 & -2 \end{vmatrix}$$

Evaluating the determinant: $\vec{n} = \hat{i}\big(1(-2) - (-2)(-6)\big) - \hat{j}\big(2(-2) - (-2)(3)\big) + \hat{k}\big(2(-6) - 1(3)\big)$ $\vec{n} = \hat{i}(-2 - 12) - \hat{j}(-4 + 6) + \hat{k}(-12 - 3)$ $\vec{n} = -14\hat{i} - 2\hat{j} - 15\hat{k}$

We can take the direction ratios of the normal to the required plane as $(a, b, c) = (14, 2, 15)$.

The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector direction ratios $(a, b, c)$ is given by: $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$

Substituting $(x_1, y_1, z_1) = (1, 1, 1)$ and $(a, b, c) = (14, 2, 15)$: $14(x - 1) + 2(y - 1) + 15(z - 1) = 0$ $14x - 14 + 2y - 2 + 15z - 15 = 0$ $14x + 2y + 15z = 31$

Thus, the correct option is C.