How to make a sum vanish? This is a very very basic question but I just cannot think of a way to tackle it for some reason. Say I have three numbers $a,b,c$ with the sum $a+b+c\neq1$. Now if I want to make this sum equal to 1 all I need to do is to make a small adjustment to these three numbers. So then $$a\rightarrow \frac{a}{a+b+c}$$
$$b\rightarrow \frac{b}{a+b+c}$$
$$c\rightarrow \frac{c}{a+b+c}$$.
Now my question is what do I do to make the sum $0$. I am sure the answer is as basic as this but my brain has gone blank. Hope someone could give me an answer thanks a lot.(Honestly I do not know what tags to use for this!)
 A: As you talked about a minimal change to your vector $(a,b,c)$ before, I'd actually suggest not looking at $a-a'+b-b'+c-c'$ as, simply put, those terms might cancel each other out. To get a minimal change in a more geometrical sense, I'd rather use some norm as $|a-a'|+|b-b'|+|c-c'|$, minimizing which is fortunately equivalent to minimizing the euclidean norm on $\mathbb{R}^3$, meaning the length of the vector $(a-a',b-b',c-c')$.
Now, if that is what we are trying to do, we actually have a fairly nice geometric interpretation of the problem. If you look at all your solutions, they are on the plane $x+y+z=0$. 
Our point $(a,b,c)$, however, is not. What we want to do is connect this point to aforementioned plane, using the shortest possible path.
Since in euclidean geometry the shortest possible path from a plane to a point is always along a straight line perpendicular to our plane, this helps us a lot - because we know what lines are normal to our plane, since it's given in normal-vector form. 
Thus we want to draw a line parallel to the normal vector $(1,1,1)$ through our point $(a,b,c)$ and find the intersection with the plane $x+y+z=0$. The result will be our point $(a',b',c')$, and the length of $(a-a',b-b',c-c')$ will be the shortest possible distance you travel in order for the sum of your components to be $0$. 
Can you take it from here? 
A: Then if $a,b,c$ are numbers that do not add up to $0$ then 
$$a\overset{\cdot bc}\longrightarrow abc$$
$$b\overset{\cdot (-2ac)}\longrightarrow -2abc$$
$$c\overset{\cdot ab}\longrightarrow abc$$ always add up to zero. But that is only one choice from infinitely many (it is just cyclically symmetric, nothing more).
