The equation of the plane passing through the | vedclass

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(A) Let the normal vectors of the two given planes be $\vec{n_1} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{n_2} = 2\hat{i} - 5\hat{j} - \hat{k}$.Sinc

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

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

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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