$\left[ {\begin{array}{*{20}{c}}
p&q&r
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
3&4&1\\
3&2&3\\
2&0&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3&0&1
\end{array}} \right]$

$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
{3p + 3q + 2r}&{4p + 2q}&{p + 3q + 2r}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3&0&1
\end{array}} \right]$

$ \Rightarrow 3p + 3q + 2r = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$

     $4p + 2q = 0 \Rightarrow q =  - 2p\,\,\,\,\,\,\,\,......\left( {ii} \right)$

      $p + 3q + 2r = 1\,\,\,\,\,\,\,\,\,\,.......\left( {iii} \right)$

On solving $(i),(ii)$ and $(iii)$ , we get 

$p=1,q=-2,r=3$

$\therefore 2p + q + r = 2\left( 1 \right) + \left( { - 2} \right) - \left( 3 \right) =  - 3$.