Of all the planes that passes through $(2,1,1)$, find the one which is furthest away from the origin. Of all the planes that passes through $P=(2,1,1)$, find the one which is furthest away from $O=(0,0,0)$. 
I know that the plane is $2x+y+z=6$, but I don’t understand why. 
I noticed that the normal vector of the plane is the one given by $P-O$ but why is that?
Wouldn’t that be finding the one which is closest?

Consider all the possible planes which pass through $P(2,1,1)$. If the plane is not perpendicular to the vector $\vec {OP}$, then it will be perpendicular at another point and this will be closer than $d=2^2+1^2+1^2=6$ units. Hence the plane will be furthest when the normal vector is $2\hat i+\hat j+\hat k$.
The equation of a plane is given by the formula $\vec r\cdot\hat n=d$
$$(x\hat i+y\hat j+z\hat k)\cdot\Biggr(\frac {2\hat i+\hat j+\hat k}{\sqrt {2^2+1^2+1^2}}\Biggr)=\sqrt {2^2+1^2+1^2}$$
$$2x+y+z=6$$
