Which of the following given value is greater than 3 12 \,?

Answer: Option D

$$\begin{array}{rl}& \text{Given, }\text{root}3\text{of}12\\ & \mathbf{Concept}\mathbf{used}:\\ & \text{If }a>b,\text{ then:}\\ & a^{\frac{1}{3}}>b^{\frac{1}{3}}\\ & \mathbf{Calculation}:\\ & \text{Option}\left(\text{A}\right):\text{root}9\text{of}1500={\left(1500\right)}^{\frac{1}{9}}\\ & \Rightarrow \text{root}3\text{of}12=12^{\frac{1}{3}}\\ & \Rightarrow \text{LCM}\left(9,3\right)=9\\ & \Rightarrow {\left(12\right)}^{\frac{1}{3}\times 9}\text{ and }{\left(1500\right)}^{\frac{1}{9}\times 9}\\ & i.e.,{\left(12\right)}^3\text{ and }{\left(1500\right)}^1\\ & i.e.,\left(1728\right)\text{ and }\left(1500\right)\\ & \text{Since, }1728>1500\\ & \Rightarrow {\left(12\right)}^{\frac{1}{3}}>{\left(1500\right)}^{\frac{1}{9}}\\ & \text{Option}\left(\text{B}\right):\text{root}5\text{of}60={\left(60\right)}^{\frac{1}{5}}\\ & \Rightarrow \text{root}3\text{of}12=12^{\frac{1}{3}}\\ & \Rightarrow \text{LCM}\left(5,3\right)=15\\ & \Rightarrow {\left(12\right)}^{\frac{1}{3}\times 15}\text{ and }{\left(60\right)}^{\frac{1}{5}\times 15}\\ & i.e.,{\left(12\right)}^5\text{ and }{\left(60\right)}^3\\ & 12^2\times 12^3\text{ and }12^3\times 5^3\\ & 144\times 12^3\text{ and }125\times 12^3\\ & \text{1}{\text{2}}^3\text{ is common both but }114>125\\ & \Rightarrow \text{Hence }12^5>60^3\text{ or }12^{\frac{1}{3}}>60^{\frac{1}{5}}\\ & \text{Option}\left(\text{C}\right):\text{root}6\text{of}121={\left(121\right)}^{\frac{1}{6}}\\ & \Rightarrow \text{root}3\text{of}12=12^{\frac{1}{3}}\\ & \Rightarrow \text{LCM}\left(6,3\right)=6\\ & \Rightarrow {\left(12\right)}^{\frac{1}{3}\times 6}\text{ and }{\left(121\right)}^{\frac{1}{6}\times 6}\\ & i.e.,{\left(12\right)}^2\text{ and }{\left(121\right)}^1\\ & i.e.,{\left(12\right)}^2\text{ and }{\left(11\right)}^2\\ & \text{Since, }12>11\\ & \Rightarrow {\left(12\right)}^{\frac{1}{3}}>{\left(121\right)}^{\frac{1}{6}}\\ & \text{Option}\left(\text{D}\right):\text{root}12\text{of}33214={\left(33214\right)}^{\frac{1}{12}}\\ & \Rightarrow \text{root}3\text{of}12=12^{\frac{1}{3}}\\ & \Rightarrow \text{LCM}\left(12,3\right)=12\\ & \Rightarrow {\left(12\right)}^{\frac{1}{3}\times 12}\text{ and }{\left(33214\right)}^{\frac{1}{12}\times 12}\\ & i.e.,{\left(12\right)}^4\text{ and }{\left(33214\right)}^1\\ & i.e.,{\left(20736\right)}^1\text{ and }{\left(33214\right)}^1\\ & \text{Since, }33214>20736\\ & \Rightarrow {\left(12\right)}^{\frac{1}{3}}>{\left(33214\right)}^{\frac{1}{12}}\\ & \therefore \text{ The required greatest value is }\text{root}12\text{of}33214.\end{array}$$