The equation of the plane passing through the | vedclass

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(C) The normal vector $\overrightarrow{n}$ of the required plane is perpendicular to the normal vectors of the given planes,$\overrightarrow{n_1} = (2

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

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(C) The normal vector $\overrightarrow{n}$ of the required plane is perpendicular to the normal vectors of the given planes,$\overrightarrow{n_1} = (2, 1, -2)$ and $\overrightarrow{n_2} = (3, -6, -2)$.Thus,$\overrightarrow{n} = \overrightarrow{n_1} 	imes \overrightarrow{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 3 & -6 & -2 \end{vmatrix}$.Calculating the determinant:$\overrightarrow{n} = \hat{i}(-2 - 12) - \hat{j}(-4 - (-6)) + \hat{k}(-12 - 3) = -14\hat{i} - 2\hat{j} - 15\hat{k}$.The equation of the plane passing through $(x_0, y_0, z_0) = (1, 1, 1)$ with normal vector $\overrightarrow{n} = (a, b, c) = (-14, -2, -15)$ is given by $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.Substituting the values:$-14(x-1) - 2(y-1) - 15(z-1) = 0$.Simplifying:$-14x + 14 - 2y + 2 - 15z + 15 = 0$.$-14x - 2y - 15z + 31 = 0$.$14x + 2y + 15z = 31$.

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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