\(L.C.M.\) 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.