Answer
$$\left( {13,\pi ,\arccos \left( {\frac{5}{{13}}} \right)} \right)$$
Work Step by Step
$$\eqalign{
& \left( {12,\pi ,5} \right) \cr
& \left( {r,\theta ,z} \right):\left( {12,\pi ,5} \right) \to r = 12,{\text{ }}\theta = \pi ,{\text{ }}z = 5 \cr
& {\text{Cylindrical to spherical }}\left( {\rho ,\theta ,\phi } \right),{\text{ }}\left( {r \geqslant 0} \right){\text{ See page 807}} \cr
& \rho = \sqrt {{r^2} + {z^2}} ,{\text{ }}\theta = \theta ,{\text{ }}\phi = \arccos \left( {\frac{z}{{\sqrt {{r^2} + {z^2}} }}} \right) \cr
& \rho = \sqrt {{{\left( {12} \right)}^2} + {{\left( 5 \right)}^2}} = \sqrt {169} = 13 \cr
& \theta = \frac{\pi }{3} \cr
& \phi = \arccos \left( {\frac{5}{{13}}} \right) \cr
& {\text{The spherical }}\left( {\rho ,\theta ,\phi } \right){\text{ coordinates are:}} \cr
& \left( {13,\pi ,\arccos \left( {\frac{5}{{13}}} \right)} \right) \cr} $$
