Why is it linearly dependent when the linear combination is zero only with none zero coefficients in 3D? Title says it all. I'm asking the geometrical sense.
I know it is linearly independent if the linear combination of vectors is zero with all the coefficients are zero. And so do dependent.

Independent $\sum a_n\mathrm{v}_n=0$, for $a_n=0$.
Dependent $\sum a_n\mathrm{v}_n=0$, not all $a_n=0$.

For mathematical notation.
I do understand in 2D, but I really don't understand why is this working in 3D.
 A: Those are not clear statements of dependence and independence of the set$\{v_1,\dots v_n\}$. 
The set would be dependent if:

There exist $\alpha_i$, not all of which are zero, such that $\sum \alpha_iv_i=0$

The set would be independent if it satisfies the negation, that there will not be such a set of nonzero coefficients:

If $\sum \alpha_iv_i=0$, then all the $\alpha_i=0$

If all the $\alpha_i=0$ then $\sum \alpha_iv_i=0$ holds all the time, so it is not interesting! It is a special case that always works. A linearly independent set is special precisely because you can't get a combination to add up to zero unless you use all zeros (which will always work.)

In any number of dimensions, linear independence expresses the idea that one vector is not in the span of the other vectors.
For example, if $\sum \alpha_iv_i=0$ where at least one of the alphas is nonzero, (say for convenience, $\alpha_1$) then $v_1=\sum_{i=2}^n \alpha_1^{-1}\alpha_iv_i$, and so $v_1$ is generatable by $v_2\dots v_n$. Then we could just throw $v_1$ out, since we know the other $v_i$ can already generate it.
So when a set is linearly independent, it means that each member really does contribute to the vector space they generate. Each element adds something new that can't be produced by the other vectors.
A: From your question, it seems you understand the concept mathematically but want to understand it geometrically.  Here's what works for me:
To briefly review the definition, we say that vectors $v_1,v_2,...,v_n$ are linearly independent if for constants $a_i$, we can only have
$
\sum_{i=1}^n a_iv_i = 0
$
when for every $i$, $a_i=0$.  
So, suppose they are linearly dependent.  Then we can say (up to a relabeling of which vector is which) that $a_1≠0$.  This allows us to say that
$$
v_1=-\sum_{i=2}^n \frac{a_i}{a_1}v_i
$$
In other words, we can put all the other vectors together tip-to-tail to get to the end of $v_1$.  In three dimensions, this would mean that we can use $v_1$, $\frac{a_2}{a_1}v_2$, and $\frac{a_3}{a_1}v_3$ to make a triangle in 3D-space.  Since the three vectors can form a triangle, we know that they all lie in the same plane.  
That is, three 3D vectors will be linearly dependent if and only if they lie in the same plane.
