$\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are on the same plane We want to show that if $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are on the same plane, then there are $A, B, C$ not all $0$ such that $A \vec a+B \vec b+C \vec c=\vec 0$.
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The solution is the following:
If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are on the same plane, then

*

*all are equal to $\overrightarrow{0}$ and then $1 \cdot \overrightarrow{a} + 1 \cdot \overrightarrow{b}+ 1 \cdot \overrightarrow{c}=\overrightarrow{0}$


*all are on the same line and one is not the $\overrightarrow{0}$
for example, $\overrightarrow{a} \neq \overrightarrow{0}$, so $\overrightarrow{b}=m \overrightarrow{a}, \overrightarrow{c}=n\overrightarrow{a}$ and then $-(m+n)\overrightarrow{a}+1\cdot \overrightarrow{b}+1 \overrightarrow{c}=\overrightarrow{0}$


*two are not the $\overrightarrow{0}$ and don't belong to the same line
for example, the $\overrightarrow{b}, \overrightarrow{c}$, then $\overrightarrow{a}=x \overrightarrow{b}+y \overrightarrow{c}$ and then $1 \cdot \overrightarrow{a}+(-x) \overrightarrow{b}+(-y) \overrightarrow{c}=\overrightarrow{0}$.
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Could you explain to me why we take these cases?
 A: They are all zero, or they aren't. If they are, done (case 1).
So, assume they aren't all zero. They are either all on the same line, or they aren't.


*

*If they are on the same line, and two, one or none of them are zero, you have something to prove (case 2).

*If they are not on the same line, and two of them are zero, then they are actually on the same line, and so we don't have to consider this case.

*Finally, if they are not on the same line, and one or none of them are zero, you have 
something to prove (case 3).
EDIT (in response to your comment below): The way you will construct your argument depends on which case you are considering. 
In the first case, we assume they are on the same line, and at least one is non-zero. 
Suppose $a$ is non-zero (since the argument is identical if we start by assuming $b$ or $c$ is the non-zero vector). If, additionally, $b$ equals zero, then $0a + 1b + 0c  =0$ and you're done (likewise if $c = 0$, then $0a + 0b + 1c = 0$).
Otherwise, if $b$ is not zero, then, because they are on the same line, there is $\alpha$ so that $a = \alpha b$ so $1 a - \alpha b = 0$.
A: Actually i would like to post a better answer (mpre intuitive) but anyway
Since the $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ vectors are on the same plane (given) either they are all zero or all on same line (which lies of course on the plane) or two of them in same line and the other on the plane not on same line.
The last case assumes implicitly that any two points (on a plane) define a unique line
So if the three points are not all co-linear, either two of them (pick any two) define a unique line and are co-linear, and the remaining third is not
