Answer:
False, counterexample below
Step-by-step explanation:
Denote the unit vectors of R^3 by [tex]e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)[/tex]. Now consider [tex]v_1=e_1, v_2=2e_1[/tex] and [tex]v_3=e_3[/tex]. We have that [tex]v_1, v_2,v_3 \in \mathbb{R}^3[/tex]. Also, the vector [tex]v_3[/tex] is not a linear combination of [tex]v_1, v_2[/tex] because any linear combination of these two vectors will have third coordinate zero, but v_3 has third coordinate 1 so they can't be equal.
However, the set [tex]\{v_1, v_2,v_3\}[/tex] is not linearly independent, because [tex]2v_1-v_2+0v_3=0[/tex] is a non-trivial linear combination of these vectors that equals zero.