If $a\leq n,b\leq n,c\leq n$ and $a+b+c\leq 2n$, then $V_a\cap V_b\cap V_c=0$ 
I did part (a). For part (b) I proved if there are subspaces $V_a,V_b,V_c$ then $a\leq n,b\leq n,c\leq n$ and $a+b+c\leq 2n$. But stuck in proving the other direction. So I suppose there are positive(or zero) integers $a,b,c$ such that $a\leq n,b\leq n,c\leq n$ and $a+b+c\leq 2n$. First I was going to prove that there is at least one pair among $\{a,b,c\}$ such that their addition $\leq n$, and then use part (a). But later I figured out that it is impossible because for example if $n=5$, and $a=b=3, c=4$ then $a+b+c=10=2n$, but non of $a+b,b+c,a+c$ is less or equals $5$. 
So, how can I prove it? Can anybody please give me at least a hint?

$\textbf{1)}$  If $a+b\le n$, then by part a) we can find subspaces $V_a, V_b$ with $V_a\cap V_b=0$, 
so then $V_a\cap V_b\cap V_c=0$ for any subspace $V_c$ with dimension $c$.
$\textbf{2)}$ If $a+b>n,\;\;$ let $B=\{v_1,\cdots, v_n\}$ be a basis for V, 
and let $V_a=\text{span}\{v_1,\cdots,v_a\}$ and let $V_b=\text{span}\{v_{n-b+1},\cdots,v_n\}$.
Since $a+b+c\le2n$, we can choose a subset $I$ of $\{1,\cdots,n-b,a+1,\cdots,n\}$ with $|I|=c$;
and then if we let $V_c=\text{span}\{v_l: l\in I\}$, $\;\;V_a\cap V_b\cap V_c=0$.
