Your work is essentially correct, although a little too detailed.
The linear dependence of 2 vectors u and v is that there exists a and b such that au+bv=0. This is what you've done, although in component form. Your result is basically that one is a scaled version of the other.
The linear dependence of 3 vectors is the same, there exist 3 constants such that au+bv+cw=0. If you want to interpret these vectors as vectors in 3 dimensional space, then yes the determinant will be zero since the 3 vectors won't fill out a solid volume. Instead they will only span a plane or line.