The set of real values of k for which the lines $x+3y+1=0,kx+2y-2=0$ and $2x-y+3=0$ form a triangle is

Lines form triangle. Therefore, $x+3y+1=0$ is not parallel to $kx+2y-2=0$ 
$\therefore \frac{-k}{2}\ne \frac{-1}{3}\Rightarrow k\ne \frac{2}{3}$ 
Also line $2x-y+3=0$  is not parallel to $kx+2y-2=0$  
$\therefore \frac{-k}{2}\ne 2\Rightarrow k\ne -4$ 
Further lines must not be concurrent
 $\therefore \left|\begin{array}{ccc}1& 3& 1\\ k& 2& -2\\ 2& -1& 3\end{array}\right|\ne 0\Rightarrow k\ne \frac{-6}{5}$