Determine the sum $L=l_1+l_2+l_3,$ as well as the absolute and

percentage errors in $L,$ given $l_1=(1\mathrm{.}72+-0\mathrm{.}04)cm,l_2=(2\mathrm{.}68+-0\mathrm{.}05)$

cm $,$ and $l_3=(0\mathrm{.}0122+-0\mathrm{.}0004)cm.$

Using the values of $l_1$ and $l_2$ from the previous question, determine

the difference

$d=l_1-l_2$ and the absolute and percentage errors in $d.$