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(A) The foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $P(-1, -1, 2)$.This point $P$ lies on the plane,and the vector $\vec{OP}

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$x+y-2z+6=0$

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(A) The foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $P(-1, -1, 2)$.This point $P$ lies on the plane,and the vector $\vec{OP} = -\hat{i} - \hat{j} + 2\hat{k}$ is normal to the plane.The equation of a plane passing through a point $\vec{a}$ with normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.Here,$\vec{a} = -\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{n} = -\hat{i} - \hat{j} + 2\hat{k}$.Substituting these values,we get: $((x+1)\hat{i} + (y+1)\hat{j} + (z-2)\hat{k}) \cdot (-\hat{i} - \hat{j} + 2\hat{k}) = 0$.$-(x+1) - (y+1) + 2(z-2) = 0$.$-x - 1 - y - 1 + 2z - 4 = 0$.$-x - y + 2z - 6 = 0$.Multiplying by $-1$,we get $x + y - 2z + 6 = 0$.

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

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