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I understand that $v_3$ is linearly dependent, however, I don't know what that means in the context of the overall question and whether or not that means $v_3$ is or isn't in the subspace.

Thanks.

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    Yes it is, since $v_3$ can be written as linear combination of $v_1, v_2$! – 2017-02-23

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It would be more correct to say that $v_1, v_2, v_3$ are linearly dependent, i.e. there exist nonzero scalars (numbers) $\alpha_1, \alpha_2, \alpha_3$ such that $$\alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3 = 0$$ = the zero vector. In other words, $v_3 = -\frac{\alpha_1}{\alpha_3} v_1 - \frac{\alpha_2}{\alpha_3} v_2$. So $v_3$ is in the subspace spanned by $v_1$ and $v_2$, because that subspace is the totality of vectors of the form $\beta_1 v_1 + \beta_2 v_2$, where $\beta_i$ are scalars.

Edit: Adren is right (some of the scalars may be zero in the definition of linear dependence). All that matters in our particular case is that $\alpha_3 \neq 0$.

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    $v_1$, $v_2$, $v_3$ are linearly dependent means that there exist scalars $\alpha_1$, $\alpha_2$, $\alpha_3$, *not all of them being zero* such that ... – 2017-02-23
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$$v_3=v_1-v_2=\alpha v_1+\beta v_2\, \text{ with } \alpha=1, \beta=-1$$ therefore by definition of $\operatorname{span}(v_1,v_2)$ we have that $v_3\in \operatorname{span}(v_1,v_2)$