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(D) The equation of a plane passing through the point $(1,2,2)$ is given by $a(x-1)+b(y-2)+c(z-2)=0$ ...$(i)$ Since the plane is perpendicular to the

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$x+y-3=0$

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(D) The equation of a plane passing through the point $(1,2,2)$ is given by $a(x-1)+b(y-2)+c(z-2)=0$ ...$(i)$ Since the plane is perpendicular to the planes $x-y+2z=3$ and $2x-2y+z+12=0$,its normal vector $(a, b, c)$ must be perpendicular to the normal vectors of the given planes,which are $\vec{n_1} = (1, -1, 2)$ and $\vec{n_2} = (2, -2, 1)$. Thus,the normal vector is $\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ 2 & -2 & 1 \end{vmatrix} = \hat{i}(-1+4) - \hat{j}(1-4) + \hat{k}(-2+2) = 3\hat{i} + 3\hat{j} + 0\hat{k}$. Comparing this with $(a, b, c)$,we get $a=3, b=3, c=0$. Substituting these values into equation $(i)$: $3(x-1) + 3(y-2) + 0(z-2) = 0$ $3(x-1 + y-2) = 0$ $x+y-3=0$.

The equation of the plane passing through the point $(1,2,2)$ and perpendicular to the planes $x-y+2z=3$ and $2x-2y+z+12=0$ is:

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