Q.Find the vector equation of a plane which is at 42 unit distance from the origin and which is normal to the vector $2\hat{i} + \hat{j} - 2\hat{k}$.

[!TLDR] A plane at a known distance $p$ from the origin, normal to a given vector, is found from the normal form $\vec r\cdot\hat n = p$: normalise the given vector first, then multiply through by $p$.

[!ANSWER] $\vec r\cdot(2\hat i+\hat j-2\hat k) = 126$

We want the plane at distance $p=42$ from the origin, normal to $\vec n = 2\hat i+\hat j-2\hat k$.

First find the unit normal: $|\vec n| = \sqrt{2^2+1^2+(-2)^2} = \sqrt{4+1+4} = \sqrt9 = 3$, so $\hat n = \dfrac{\vec n}{|\vec n|} = \dfrac{2\hat i+\hat j-2\hat k}{3}$.

By the normal form of the equation of a plane, $\vec r\cdot\hat n = p$:

Multiplying both sides by $3$:

[!ANSWER] $\vec r\cdot(2\hat i+\hat j-2\hat k) = 126$

1. Compute $|\vec n|$ for the given normal vector.
2. Form the unit vector $\hat n = \vec n/|\vec n|$.
3. Apply the normal form $\vec r\cdot\hat n = p$ with the given distance $p$.
4. Clear the fraction by multiplying both sides by $|\vec n|$ to present the equation with the original (non-unit) normal vector.

Writing $\vec r\cdot(2\hat i+\hat j-2\hat k) = 42$ directly, i.e. using the un-normalised vector with the raw distance $p$ instead of first dividing by $|\vec n|=3$ (this only works when $|\vec n|=1$ already). Also, dropping the middle term's sign and writing the normal as $2\hat i - \hat j - 2\hat k$ instead of the printed $2\hat i+\hat j-2\hat k$.