Answer
$$\left( {\frac{{25\sqrt 2 }}{2}, - \frac{\pi }{4}, - \frac{{25\sqrt 2 }}{2}} \right)$$
Work Step by Step
$$\eqalign{
& \left( {25, - \frac{\pi }{4},\frac{{3\pi }}{4}} \right) \cr
& {\text{spherical}}\left( {\rho ,\theta ,\phi } \right):\left( {25, - \frac{\pi }{4},\frac{{3\pi }}{4}} \right) \to \rho = 25,{\text{ }}\theta = - \frac{\pi }{4},{\text{ }}\phi = \frac{{3\pi }}{4} \cr
& {\text{Spherical to cylindrical }}\left( {r,\theta ,z} \right),{\text{ }}\left( {r \geqslant 0} \right){\text{ See page 807}} \cr
& \cr
& {r^2} = {\rho ^2}{\sin ^2}\phi ,{\text{ }}\theta = \theta ,{\text{ }}z = \rho \cos \phi \cr
& {r^2} = {\left( {25} \right)^2}{\sin ^2}\left( {\frac{{3\pi }}{4}} \right) = \frac{{625}}{2} \to r = \frac{{25\sqrt 2 }}{2} \cr
& \theta = - \frac{\pi }{4} \cr
& z = 25cos\left( {\frac{{3\pi }}{4}} \right) = - \frac{{25\sqrt 2 }}{2} \cr
& \cr
& {\text{The cylindrical }}\left( {r,\theta ,z} \right){\text{ coordinates are:}} \cr
& \left( {\frac{{25\sqrt 2 }}{2}, - \frac{\pi }{4}, - \frac{{25\sqrt 2 }}{2}} \right) \cr} $$
