Showing this set in linearly dependent 
Being $V$ a linear space of $\Bbb K$. Show that if $v_1, v_2, v_3$ are vectors of $V$ and $_1, _2, _3 ∈ $ then the set $${_2v_3 − _3v_2, _1v_2 − _2v_1, _3v_1 − _1v_3}$$
  is linearly dependent.

My attempt:
If the set is linearly dependent then there are scalars $\beta1$ $\beta2$ and $\beta3$ not all zero that:
$$\beta_1(_2v_3 − _3v_2) + \beta_2(_1v_2 − _2v_1) + \beta_3(_3v_1 − _1v_3) = 0 $$
I started solving the equation and I decided to isolate $v_3$, so that I can write that I can express $v_3$ as a linear combination of $v_2$ and $v_1$. But for that $\beta_1\alpha_2 + \beta_3\alpha_1$ might be different that zero $\dots$ 
But how can I justify that?
 A: My thought: Can I write the thirs vector as a linear combination of the first two? Where to start: The scalar of $v_3$ in the third vector is $α_3$ but in the first vector it is $α_2$. So let's multiply the first vector by $\frac{α_3}{α_2}$ assuming that $α_2\neq 0$. Then repeat for $v_1$ and the second vector and see if it works (which it does): $$α_3v_1-α_1v_3=-\frac{α_1}{α_2}(α_2v_3-α_3v_2)-\frac{α_3}{α_2}(α_1v_2-α_2v_1)$$ if of course $α_2\neq 0$. Else...
A: Since it's possible that $\vec v_1, \vec v_2, \vec v_3$ are linearly independent vectors, we want to solve for $\beta_1, \beta_2, \beta_3$ in terms of $\alpha_1, \alpha_2, \alpha_3$ subject to the constraints that:
$$\begin{cases}
\alpha_3\beta_3 - \alpha_2\beta_2 = 0 \\
\alpha_1\beta_2 - \alpha_3\beta_1 = 0 \\
\alpha_2\beta_1 - \alpha_1\beta_3 = 0
\end{cases}$$
By inspection, note that we can choose:
$$\begin{cases}
\beta_1 = \alpha_1 \\
\beta_2 = \alpha_3 \\
\beta_3 = \alpha_2
\end{cases}$$
A: Multiply the first by $a_1$ then you get $a_2a_1v_3-a_3a_1v_2$. Multiply the second by $a_3$ then you get $a_1a_3v_2-a_2a_3v_1$. Summing you obtain $a_2(a_1v_3-a_3v_1)$. If $a_2=0$ the set is trivially dependend because the first vector is $a_1v_2$ and the second is $a_3v_2$. If not, then
$$\frac{a_1}{a_2}(a_2v_3-a_3v_2)+\frac{a_3}{a_2}(a_1v_2-a_2v_1)=a_1v_3-a_3v_1$$
