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(C) The equation of a plane passing through $(1, 1, 1)$ is given by $a(x - 1) + b(y - 1) + c(z - 1) = 0$. Since this plane is perpendicular to the pla

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$14x + 2y + 15z = 31$

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(C) The equation of a plane passing through $(1, 1, 1)$ is given by $a(x - 1) + b(y - 1) + c(z - 1) = 0$. Since this plane is perpendicular to the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$,its normal vector $\vec{n} = (a, b, c)$ must be perpendicular to the normal vectors of the given planes,$\vec{n_1} = (2, 1, -2)$ and $\vec{n_2} = (3, -6, -2)$. Thus,the normal vector $\vec{n}$ is given by the cross product $\vec{n_1} 	imes \vec{n_2}$: $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 3 & -6 & -2 \end{vmatrix} = \hat{i}(-2 - 12) - \hat{j}(-4 + 6) + \hat{k}(-12 - 3) = -14\hat{i} - 2\hat{j} - 15\hat{k}$. Taking the normal vector as $(14, 2, 15)$,the equation of the plane is $14(x - 1) + 2(y - 1) + 15(z - 1) = 0$. Expanding this,we get $14x - 14 + 2y - 2 + 15z - 15 = 0$,which simplifies to $14x + 2y + 15z = 31$.

The equation of the plane,passing through the point $(1, 1, 1)$ and perpendicular to the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$,is

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