$C_{M_{HCl}}=10^{-2}=0,01M$
$\Rightarrow n_{HCl}=0,3.0,01=0,003 mol$
$pOH=14-12=2$
$\Rightarrow C_{M_{NaOH}}=10^{-2}=0,01M$
$\Rightarrow n_{NaOH}=0,01.0,2=0,002 mol$
$NaOH+HCl\to NaCl+H_2O$
$\Rightarrow n_{HCl\text{dư}}=n_{H^+}=0,003-0,002=0,001mol$
$V_{dd}=0,2+0,3+0,5=1l$
$\Rightarrow [H^+]=0,001M$
$pH=-lg0,001=3$