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

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$\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 15$

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(B) The vector equation of a plane at a distance $d$ from the origin and normal to a vector $\vec{n}$ is given by $\vec{r} \cdot \hat{n} = d$, where $\hat{n} = \frac{\vec{n}}{|\vec{n}|}$ is the unit normal vector.Given $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $d = 5$.Calculate the magnitude of $\vec{n}$: $|\vec{n}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.The unit normal vector is $\hat{n} = \frac{2\hat{i} + \hat{j} - 2\hat{k}}{3}$.The equation of the plane is $\vec{r} \cdot \left( \frac{2\hat{i} + \hat{j} - 2\hat{k}}{3} \right) = 5$.Multiplying both sides by $3$, we get $\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 15$.

The vector equation of a plane which is at a distance of $5 \text{ units}$ from the origin and normal to the vector $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$ is:

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