Ok, I'll do your homework. First: $$c ={1\over 3}(k+l+m)$$
$\vec{CA} = {4\over 7}\vec{CN}$ so $7a-7c =4n-4c$ and thus $$7a =3c+4n \Longrightarrow a= {1\over 7}(k+l+m+4n)$$
so finally we have:
\begin{eqnarray}\vec{AL} + \vec{AM} + 4\vec{AN} &=& l+m+4n-6a \\\&=& {1\over 7}(7l+7m+28n-6k-6l-6m-24n)\\\ &=& {1\over 7}(l+m+4n-6k)\\\ &=& {1\over 7}(l+m+4n+k)-k \\\ &=& a-k \\\ &=&-\vec{AK} \end{eqnarray}