A plane ax+by+cz+1=0 is perpendicular to the | vedclass

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(D) The normal vectors to the planes $2x-2y+z=0$ and $x-y+2z=4$ are $\vec{n_1} = (2, -2, 1)$ and $\vec{n_2} = (1, -1, 2)$ respectively.The normal vect

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$2$

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(D) The normal vectors to the planes $2x-2y+z=0$ and $x-y+2z=4$ are $\vec{n_1} = (2, -2, 1)$ and $\vec{n_2} = (1, -1, 2)$ respectively.The normal vector to the plane $ax+by+cz+1=0$ is $\vec{n} = (a, b, c)$.Since the plane is perpendicular to the given two planes, its normal vector $\vec{n}$ must be parallel to the cross product of $\vec{n_1}$ and $\vec{n_2}$.$\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 1 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4+1) - \hat{j}(4-1) + \hat{k}(-2+2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.Thus, the normal vector is proportional to $(-3, -3, 0)$, which simplifies to $(1, 1, 0)$.The equation of the plane is $1(x-1) + 1(y+2) + 0(z-1) = 0$, which simplifies to $x+y+1=0$.Comparing this with $ax+by+cz+1=0$, we get $a=1, b=1, c=0$.Therefore, $a+b-c = 1+1-0 = 2$.

$A$ plane $ax+by+cz+1=0$ is perpendicular to the two planes $2x-2y+z=0$ and $x-y+2z=4$ and passes through the point $(1, -2, 1)$. Then $a+b-c=$

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