The equation of the plane passing through the | vedclass

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(C) The normal vector $\vec{n}$ to the required plane is the cross product of the normals of the two given planes,$\vec{n_1} = (3, 1, -2)$ and $\vec{n

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$11x + y + 17z + 38 = 0$

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(C) The normal vector $\vec{n}$ to the required plane is the cross product of the normals of the two given planes,$\vec{n_1} = (3, 1, -2)$ and $\vec{n_2} = (2, -5, -1)$.$\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & -2 \ 2 & -5 & -1 \end{vmatrix} = \hat{i}(-1 - 10) - \hat{j}(-3 - (-4)) + \hat{k}(-15 - 2) = -11\hat{i} - \hat{j} - 17\hat{k}$.We can take the normal vector as $\vec{n} = (11, 1, 17)$.The equation of the plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.Substituting the point $(1, 2, -3)$ and the normal vector $(11, 1, 17)$:$11(x - 1) + 1(y - 2) + 17(z + 3) = 0$$11x - 11 + y - 2 + 17z + 51 = 0$$11x + y + 17z + 38 = 0$.

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

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