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?
 A: 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 $\vec r\cdot\hat n=d$
$$(x\hat i+y\hat j+z\hat k)\cdot\Biggr(\frac {2\hat i+\hat j+\hat k}{\sqrt {2^2+1^2+1^2}}\Biggr)=\sqrt {2^2+1^2+1^2}$$
$$2x+y+z=6$$
A: $\vec n \cdot (\vec r - (2,1,1))=0$.
Choose the normal : $\vec n =(2,1,1)$;
($\vec n$ is direction vector joining $(0,0,0)$ to $(2,1,1)$).
Then 
$(2,1,1)\cdot (x-2,y-1,z-1)=0$;
$2(x-2)+(y-1)+ (z-1)=0$;
$2x+y+z-6=0$.
Reasoning:
Consider a sphere centered at origin with 
radius  $R=\sqrt{2^2+1^1+1^1}$ passing through $(2,1,1)$.
With the above choice of $\vec n$  the plane lies in the tangential plane of the sphere at $(2,1,1)$, i.e. touches  the sphere.
Any other choice  of $\vec n$ will tilt the plane and have it intersect the sphere in other points, means: there are points on the plane within the sphere, smaller distance.
A: Try to visualise the problem in two dimensions first: given a point $P\in\Bbb R^2$, what is the line through $P$ that is furthest from the origin? You ought to be able to convince yourself that this line is perpendicular to $OP$. Draw a picture if it helps.
A: 
In right angle triangle OQP right angle at Q . Hypotonous is OP is the longest side i.e $OP\ge OQ$ hence when distance is OP plane is at maximum distance.
