Answer
$$\left( {\sqrt 6 ,{{\tan }^{ - 1}}\left( 2 \right),\arccos \left( {\frac{1}{{\sqrt 6 }}} \right)} \right)$$
Work Step by Step
$$\eqalign{
& \left( { - 1,2,1} \right) \cr
& \left( {x,y,z} \right):{\text{ }}\left( { - 1,2,1} \right) \to x = - 1,{\text{ }}y = 2,{\text{ }}z = 1 \cr
& {\text{Rectangular to spherical }}\left( {\rho ,\theta ,\phi } \right) \cr
& {\rho ^2} = {x^2} + {y^2} + {z^2} \to \rho = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt 6 \cr
& \tan \theta = \frac{y}{x} \to \theta = {\tan ^{ - 1}}\left( {\frac{2}{{ - 1}}} \right) = {\tan ^{ - 1}}\left( 2 \right) + \pi \cr
& \phi = \arccos \left( {\frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) \to \phi = \arccos \left( {\frac{1}{{\sqrt 6 }}} \right) \cr
& {\text{The spherical }}\left( {\rho ,\theta ,\phi } \right){\text{ coordinates are:}} \cr
& \left( {\sqrt 6 ,{{\tan }^{ - 1}}\left( 2 \right),\arccos \left( {\frac{1}{{\sqrt 6 }}} \right)} \right) \cr} $$
