Let two planes be P_1: 2x - y + z = 2 and P_2 | vedclass

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(C) The normal vector $\vec{n_1}$ of plane $P_1$ is $\langle 2, -1, 1 \rangle$ and the normal vector $\vec{n_2}$ of plane $P_2$ is $\langle 1, 2, -1 \

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$x - 3y - 5z + 20 = 0$

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(C) The normal vector $\vec{n_1}$ of plane $P_1$ is $\langle 2, -1, 1 angle$ and the normal vector $\vec{n_2}$ of plane $P_2$ is $\langle 1, 2, -1 angle$.Since the required plane is perpendicular to both $P_1$ and $P_2$,its normal vector $\vec{n}$ must be parallel to $\vec{n_1} 	imes \vec{n_2}$.$\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 1 \ 1 & 2 & -1 \end{vmatrix} = \hat{i}(1 - 2) - \hat{j}(-2 - 1) + \hat{k}(4 + 1) = -\hat{i} + 3\hat{j} + 5\hat{k}$.Thus,the normal vector is $\langle -1, 3, 5 angle$.The equation of the plane passing through $(-1, 3, 2)$ with normal $\langle -1, 3, 5 angle$ is given by:$-1(x - (-1)) + 3(y - 3) + 5(z - 2) = 0$$-1(x + 1) + 3(y - 3) + 5(z - 2) = 0$$-x - 1 + 3y - 9 + 5z - 10 = 0$$-x + 3y + 5z - 20 = 0$Multiplying by $-1$,we get $x - 3y - 5z + 20 = 0$.

Let two planes be $P_1: 2x - y + z = 2$ and $P_2: x + 2y - z = 3$. Find the equation of the plane passing through the point $(-1, 3, 2)$ and perpendicular to both planes $P_1$ and $P_2$.

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