Sufficiently long long division

$\LARGE{ \begin{array}{rlll} \phantom{0}\ \boxed{\color{#D61F06}{6}} \ \boxed{\color{#D61F06}{7}} \ \boxed{\color{#D61F06}{8}} && \\[-4pt] \boxed{\color{#D61F06}{1}} \ \boxed{\color{#D61F06}{1}}\ \enclose{longdiv}{\boxed{\color{#D61F06}{7}} \ \boxed{\color{#D61F06}{4}} \ \boxed{\color{#D61F06}{6}} \ \boxed{\color{#D61F06}{7}} } && \\[-4pt] \underline{\boxed6\ \boxed{\color{#D61F06}{6}}\, \phantom0 \phantom0 \phantom0 \phantom0} && \\[-4pt] \boxed{\color{#D61F06}{8}} \ \boxed{\color{#D61F06}{6}} \phantom0 \, \, \phantom0 && \\[-3pt] \underline{\boxed{7} \ \boxed{\color{#D61F06}{7}} \,\, \phantom0 \phantom0 } && \\[-3pt] \boxed{\color{#D61F06}{9}} \ \boxed{\color{#D61F06}{7}} && \\[-2pt] \underline{\boxed{8} \ \boxed{\color{#D61F06}{8}}} && \\[-2pt] \boxed{9} \end{array} }$ $\small\color{forestgreen}{\text{Hint:The choice for divisor is between 11 and 12.}\\\text{For 12 second block of third row is turning to 0}\\\text{Hence divisor must be 11}}$