How do we know if a third vector is on the plane of the first and second vector? I have to add its components and if it gives me zero,then is it on the plane?
Suppose that the vectors have three components.
Thank you.

Other answers are right, but I personally avoid the use of determinant when I can, so here is an alternative solution.

I have to add its components and if it gived me zero,then is it on the plane?

I don't really get what you mean with that. 
Let's call your vectors $u$,$v$ and $w$. Let's say that $u$ and $v$ are your first vectors spanning a plane, and you're asking if $w$ is in that plane.
By "adding it's components", are your adding all the components of $w$ (i.e $w_1+w_2+w_3$)? That doesn't really make sense in this context.
However, if $w$ is in this plan, it is a linear combinaison of $u$ and $v$. So, it can be written in the form of: $$\alpha u + \beta v = w$$
Hence, you have to solve the system:
$$\begin{cases}
\alpha u_1+\beta v_1=w_1 \\
\alpha u_2+\beta v_2=w_2 \\
\alpha u_3+\beta v_3=w_3 \\
\end{cases}$$
You then have three possibilities:


*

*If you find a uniq solution for $\alpha,\beta$, all vectors are in the same plane. 

*If you find infinite solutions, that means that all vectors are colinear, and $u$ and $v$ did't span a plane in the first place. 

*If you don't find any solution, the vectors are linearly independant, and therefore, $w$ isn't in the plane spanned by $u$ and $v$
