Did I go about determining the coplanarity of these three vectors wrong? I asked this question a few days ago, where the question was this:

I have a task stating this:

Determine if the following vectors are coplanar.
    Assume that $v_1$, $v_2$ and $v_3$ are not coplanar.
$w_1=4\vec v_1+3\vec v_2$
$w_2=\vec v_2+4\vec v_3$
$w_3=-\vec v_1-3\vec v_3$

I don't quite understand how I'll do this when I do not know the values of any of the vectors. Also, what significance does the information "$v_1$, $v_2$ and $v_3$ are not coplanar" have in terms of the solution? I'm guessing knowing that helps decide whether they're coplanar or not, but I can't see how.

In that question we arrived at the conclusion that they are coplanar. However, since then, I've attempted to find a way to mathmatically prove that they are coplanar, but instead I seem to arrive at the conclusion that they are in fact NOT coplanar. What I've done looks like this:
If the sum of the three vectors equals the zero vector in any way other than by multiplying every vector by zero, then they are coplanar. So if I can prove that one of the vectors can be represented as a multiple of the other, then they are coplanar. So to try and prove this, I’ll try to add two vectors.
$$\vec w_1+\vec w_2=(4\vec v_1+3\vec v_2)+(\vec v_2+4\vec v_3)$$
From there, I realized I could do this instead:
$$\vec w_1-3\vec w_2=(4\vec v_1+3\vec v_2)-3(\vec v_2+4\vec v_3)$$
$$\vec w_1-3\vec w_2=4\vec v_1+3\vec v_2-3\vec v_2-12\vec v_3$$
$$\vec w_1-3\vec w_2=4\vec v_1-12\vec v_3$$
Now the two vectors are written in the same terms as the third equation. However, this poses the problem that I need to something to multiply $(4-12)$ with to make it $(-1-3)$. Which simply isn't possible, meaning there is no way I can represent $\vec w_1$ and $\vec w_2$ as a linear combination of $\vec w_3$. This again ultimately means the vectors are linearly independent, and thus also not coplanar.
So, did I do something wrong here? Am I trying to prove this the wrong way, or was the answer for the other question simply wrong? As far as I can tell now, they are not coplanar, but in the other question we concluded that they were.
 A: You did the right thing in general, but trying out specific cases is a bad way to go -- you might miss the one combination of the $w$s that actually comes out to zero. 
Instead, what you want to do is say this:
Let's suppose that the $w$ are coplanar. Then there are numbers $a$, $b$, and $c$ such that
$$ 
aw_1 + b w_2 + c w_3 = 0
$$
Now I can substitute in the expressions for the $w$s to get that 
$$ 
a(4v_1 + 3v_2)  + b (v_2 + 4v_3) + c (-v_1 - 3 v_3) = 0
$$
Expanding, you get
$$
4a v_1 + 3a v_2 + bv_2 + 4b v_3 - c v_1 -3c v_3 = 0 \\
(4a  - c) v_1 + (3a + b) v_2 + (4b -3c) v_3 = 0 \\
$$
So IF you could find a (not all zeros) combination of the $w$s that's zero, you'd also have a combination of the $v$s that's zero. You might think that'd mean that the $v$s are coplanar, but it only means that if the coefficients aren't all zero. 
What would make all three of those coefficients be zero? Let's do some algebra. We'll assume that 
$$
4a - c = 0 \\
3a + b = 0 \\
4b - 3c = 0
$$
Multiply the first equation by 3 to get
$$
12a - 3c = 0 \\
3a + b = 0 \\
4b - 3c = 0
$$
Subtract the thrid equation from the first to get
$$
12a  - 4b = 0 \\
3a + b = 0 \\
4b - 3c = 0
$$
Multiply the second equation by $4$ to get
$$
12a  - 4b = 0 \\
12a + 4b = 0 \\
4b - 3c = 0
$$
Subtract the first from the second to get
$$
12a  - 4b = 0 \\
8b = 0 \\
4b - 3c = 0
$$
So $b = 0$, which (from the first equation) says that $a = 0$, which from the third equation tells us $c = 0$. 
In summary: if the three $w$s were coplanar, we'd have found $a,b,c$, not all zero, with $aw_1 + bw_2 + cw_3 = 0$. That would in turn make a combination of the $v$s be zero. That latter combination has all-zero coeffs only if the former does, so you'd have found a not-all-zero combination of the $v$s that's zero, which would imply the $v$s are coplanar. Since they're not, we've got a contradiction. The assumption that the $w$s are coplanar must be wrong. Hence they're non-coplanar. 
A: I think you are correct: if one is a linear combination of the other two, it can be any one. You picked $\vec{w}_3$ and showed that it is not.
