Let u,v,w be three linearly independent vectors in a 7x1 matrix, determine a value of k 
Let $u,v,w$ be three linearly independent vectors in $\mathbb{R}^7$ , determine a value of $k$
$k =$ ________ , so that the set $S = {u -5v, v-3w, w-ku} $ is linearly dependent

I'm very confused, and not sure how to even a start this one to solve
 A: Assuming $ u $ to be non-zero, we perform the following quick calculation:
$$ (u-5v) + 5(v-3w) + 15(w-ku) = 0 $$( for linear dependence).
Then we easily obtain $ (1-15k) = 0$ which gives us a value of k to be $\frac{1}{15}$
A: Consider an arbitrary linear combination of $(u-5v),(v-3w),(w-ku)$ which equals zero.
$\alpha_1(u-5v)+\alpha_2(v-3w)+\alpha_3(w-ku) = 0$
Distributing and reorganizing terms:
$(\alpha_1-\alpha_3k)u + (\alpha_2-5\alpha_1)v+(\alpha_3-3\alpha_2)w=0$
We know, however that $u,v,w$ are linearly independent.  They only way the above line can be true then is if:

$\alpha_1-\alpha_3k = \alpha_2-5\alpha_1=\alpha_3-3\alpha_2=0$

Continuing to manipulate the expressions, we try to find a value of $k$ such that at least one of $\alpha_1,\alpha_2,\alpha_3$ are nonzero which would thus make the above vectors linearly dependent.

$\alpha_2=5\alpha_1\Rightarrow a_3=15\alpha_1\Rightarrow a_1=15k\alpha_1$

This final conclusion is true when either:

$\alpha_1=0$ or when $k=\frac{1}{15}$.  But if $\alpha_1=0$ then $\alpha_2$ and $\alpha_3$ must also be zero implying the vectors are still linearly independent.

This implies:

The vectors are linearly dependent if and only if $k=\frac{1}{15}$

