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(B) The vector equation of a plane at a distance $d$ from the origin and normal to the vector $\vec{n}$ is given by $r \cdot \hat{n} = d$,where $\hat{

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$r \cdot (2\hat{i} + \hat{j} + 2\hat{k}) = 24$

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(B) The vector equation of a plane at a distance $d$ from the origin and normal to the vector $\vec{n}$ is given by $r \cdot \hat{n} = d$,where $\hat{n}$ is the unit normal vector.Given $d = 8$ and $\vec{n} = 2\hat{i} + \hat{j} + 2\hat{k}$.First,calculate the magnitude of $\vec{n}$:$|\vec{n}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.Now,find the unit normal vector $\hat{n}$:$\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3}$.The equation of the plane is $r \cdot \hat{n} = d$:$r \cdot \left( \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3} \right) = 8$.Multiplying both sides by $3$,we get:$r \cdot (2\hat{i} + \hat{j} + 2\hat{k}) = 24$.

The vector equation of a plane,which is at a distance of $8$ units from the origin and which is normal to the vector $\vec{n} = 2\hat{i} + \hat{j} + 2\hat{k}$,is

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