How do I find out of these vectors are coplanar? 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.
 A: Hint: in a 3-dimensional space, $v_1, v_2, v_3$ being not coplanar is equivalent to saying that they are linearly independent, i.e., they form a basis. The question is to show that the same holds (or doesn't hold) for $w_1, w_2, w_3$.
A: If we set $$A=\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}\in M_{3\times 3}(F)$$ then $\det(A)$ is the same as $\det(B)$ where $$B=\begin{pmatrix} 4v_1+3v_2 \\ v_2+4v_3 \\ -v_3-v_1 \end{pmatrix}$$ since the Elementary row operations let us for that.
A: You have to set up a system of linear equations and show that there is a non trivial solution. You can start like a*w1 + b*w2 + c* w3 = 0, then substitute the wi with your equations. Then, you get equations in v1,v2,v3 and can find out, whether there is a nontrivial solution by applying the knowledge that v1,v2,v3 are not coplanar: 
Solve a*w1+b*w2+c*w3 = 0 for a,b,c: 
a (4v1+3v2)+b(v2+4v3)+c(-v1-3v3) = 0 
yields: 
v1 (4a-c) + v2 (3a+b) + v3 ( 4b-3c) = 0
and since v1,v2,v3 are not coplanar: 
4a-c = 0 and 3a+b = 0 and 4b-3c = 0
which leads to a = b = c = 0 (c=4a, b = -3a, therefore 0 = 4b-3c = -12a-12a = -24a => a =0; => b =0 => c=0)
Therefore they are not coplanar. 
Sorry for the formatting.
