We are calculating
$$\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)$$
We have to start by making the demoninators the same
$$\left({\frac{{\mathtt{9}}}{{\mathtt{12}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{8}}}{{\mathtt{12}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{6}}}{{\mathtt{12}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{4}}}{{\mathtt{12}}}}\right)$$
$$\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{10}}}{{\mathtt{12}}}}\right)$$
$${\frac{{\mathtt{11}}}{{\mathtt{12}}}}$$
This answer can't be reduced at all, this is the simplest term. 11 is the numerator and 12 is the demoninator, so the demoninator of the resulting fraction is 12