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(A) The normal vector $\vec{n}$ to the plane is the vector from the origin $O(0, 0, 0)$ to the foot of the perpendicular $M(-1, -2, 2)$.Thus,$\vec{n}

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$\bar{r} \cdot(-\hat{i}-2 \hat{j}+2 \hat{k})=9$

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(A) The normal vector $\vec{n}$ to the plane is the vector from the origin $O(0, 0, 0)$ to the foot of the perpendicular $M(-1, -2, 2)$.Thus,$\vec{n} = \vec{OM} = -\hat{i} - 2\hat{j} + 2\hat{k}$.The equation of a plane passing through a point $M$ with normal vector $\vec{n}$ is given by $\vec{r} \cdot \vec{n} = \vec{OM} \cdot \vec{n}$.Here,$\vec{OM} \cdot \vec{n} = (-1, -2, 2) \cdot (-1, -2, 2) = (-1)^2 + (-2)^2 + (2)^2 = 1 + 4 + 4 = 9$.Therefore,the vector equation of the plane is $\vec{r} \cdot(-\hat{i} - 2\hat{j} + 2\hat{k}) = 9$.

If the foot of the perpendicular drawn from the origin to a plane is $M(-1, -2, 2)$,then the vector equation of the plane is

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