If the foot of the perpendicular from O(0, 0, 0) to a plane is P (1, 2, 2). Then, the equation of the plane is 

$x + 2y + 2z - 9 = 0 $

The plane passes through P(1, 2, 2) and is normal to $\vec{OP}= \hat{i} + 2\hat{j} + 2\hat{k}$. So, its vector equation is

$\vec{r}. \vec{OP}= (\hat{i} + 2\hat{j} + 2\hat{k}). \vec{OP}$         [Using : $\vec{r}. \vec{n} = \vec{a}.\vec{n}]$

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

$⇒ \vec{r}.(\vec{i}+2\hat{j} + 2\hat{k})= 9 ⇒ x + 2y + 2z = 9$