Answer
See explanation
Work Step by Step
Because $v_{4}$ is not in Span $\left\{v_{1}, v_{2}, v_{3}\right\},$ then $v_{4}$ is not a linear combination of vectors $v_{1}, v_{2}, v_{3}$
Span $\left\{v_{1}, v_{2}, v_{3}\right\}$ is a three dimensional subspace of $\mathbb{R}^{5},$ so $v_{4}$ "exists" in $\mathbb{R}^{5} \backslash \operatorname{Span}\left\{v_{1}, v_{2}, v_{3}\right\}$
So, $\left\{v_{1}, v_{2}, v_{3}, v_{4}\right\}$ must be linearly independent.
