jmc

Since $1_4+3_4+2_4=12_4$, we carry over the $1$. Then we have $1_4+1_4+2_4+3_4=13_4$, so we carry over another $1$. For the leftmost column, we have $1_4+1_4+3_4+1_4=12_4$. In column format, this reads \begin{array}{c@{}c@{\;}c@{}c@{}c@{}c} & & & _{1} &_{1}&\\ & & & 1& 1 & 1_4\\ & & & 3 & 2 & 3_4\\& & + & 1 & 3 & 2_4\\ \cline{2-6} & & 1 & 2 & 3& 2_4\\ \end{array}The sum is $\boxed{1232_4}$.