The vector equation of the plane which is at a distance of 6 units from the origin and has2, -1, 2 as the direction ratios of normal to it, is :

$\vec{r}. (2\hat{i}-\hat{j}+2\hat{k})= 18 $

The correct answer is Option (1) → $\vec{r}. (2\hat{i}-\hat{j}+2\hat{k})= 18 $

$\vec n=2\hat i-\hat j+2\hat k$

$\hat n=\frac{\vec n}{|\vec n|}=\frac{2\hat i-\hat j+2\hat k}{\sqrt{2^2+(-1)^2+2^2}}=\frac{1}{3}(2\hat i-\hat j+2\hat k)$

$\vec r.\hat n=d$

$⇒\vec r.\frac{(2\hat i-\hat j+2\hat k)}{3}=6$

$\vec r.(2\hat i-\hat j+2\hat k)=18$