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(B) The required plane passes through the point $(1,1,1)$ and is perpendicular to the planes $2x-y-2z=5$ and $3x-6y+2z=7$. The normal vectors of these

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$14x+10y+9z=33$

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(B) The required plane passes through the point $(1,1,1)$ and is perpendicular to the planes $2x-y-2z=5$ and $3x-6y+2z=7$. The normal vectors of these planes are $\vec{n_1} = 2\hat{i} - \hat{j} - 2\hat{k}$ and $\vec{n_2} = 3\hat{i} - 6\hat{j} + 2\hat{k}$. The normal vector $\vec{n}$ of the required plane is $\vec{n_1} 	imes \vec{n_2}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & -2 \ 3 & -6 & 2 \end{vmatrix} = \hat{i}(-2-12) - \hat{j}(4+6) + \hat{k}(-12+3) = -14\hat{i} - 10\hat{j} - 9\hat{k}$.The equation of the plane is $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.$-14(x-1) - 10(y-1) - 9(z-1) = 0$.$-14x + 14 - 10y + 10 - 9z + 9 = 0$.$-14x - 10y - 9z + 33 = 0$.$14x + 10y + 9z = 33$ is incorrect in the provided options,let's recheck the determinant calculation.$\begin{vmatrix} x-1 & y-1 & z-1 \ 2 & -1 & -2 \ 3 & -6 & 2 \end{vmatrix} = (x-1)(-2-12) - (y-1)(4+6) + (z-1)(-12+3) = -14(x-1) - 10(y-1) - 9(z-1) = -14x + 14 - 10y + 10 - 9z + 9 = -14x - 10y - 9z + 33 = 0$.Thus,$14x + 10y + 9z = 33$.

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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