If we take a=0 then $v_1$ will become zero vector which in result will make {$v_1$, $v_2$, $v_3$} linearly dependent as we can then represent, v1 = 0.v2 + 0.v3 . Hence a$\neq$0.
If we take b=1 then $v_2$ and $v_3$ will become same vector. Hence b$\neq$1 for {$v_1$, $v_2$, $v_3$} to be linearly independent.
$\therefore$ b is correct.