We want the lines to be parallel and not the same line in order to have 0 solutions. Parallel implies the slopes of the lines are equal, so we have $-\frac{8}{6 k}=-\frac{k}{12}$. Cross multiplying, we get $96=6 k^{2}$, so $k^{2}=$ 16 and our positive solution is then $k=4$. 
For what positive value of $k$ does the following system of equations have no solutions?~\uline{~~~~~~~~~~}~  $$ \begin{cases} 8 x+6 k y=17 \textbackslash\textbackslash{} k x+12 y=16 \end{cases}$$ 

  $$2$$ 
  $\frac{56}{15}$ 
  $$4$$ 
  $$8$$ 