(a) $\root 3 \of 9 ,\,\,\root 4 \of {11} ,\,\,\root 6 \of {17} $

$L.C.M.$ of $3, 4, 6$ is $12$

$\therefore \sqrt[3]{9} = {9^{1/3}} = {({9^4})^{1/12}} = {(6561)^{1/12}}$

$\root 4 \of {11} = {(11)^{1/4}}{({11^3})^{1/12}} = {(1331)^{1/12}}$,

$\root 6 \of {17} = {(17)^{1/6}} = {({17^2})^{1/2}} = {(289)^{1/12}}$

Hence, $\root 3 \of 9 $ is the greatest number.