Answer
False.
Work Step by Step
Let $\vec{v}_{1}=(1,1,1,1)$, $\vec{v}_{2}=(2,2,2,2)$, $\vec{v}_{3}=(1,2,3,4)$, and $\vec{v}_{4}=(3,3,3,3)$. Clearly, the vectors are linearly dependent, since $\vec{v}_{1}$, $\vec{v}_{2}$, and $\vec{v}_{4}$ are all scalar multiples of one another, but $\vec{v}_{3}$ is not a linear combination of any of the other three.
Recall the warning beneath Theorem 7: we only need one of the vectors to be a linear combination of others in the set for the set to be linearly dependent. It need not be that every vector in the set is a linear combination of other vectors in the set.
