Answer
$$\left( {\frac{7}{{\sqrt 2 }},\frac{\pi }{4}, - \frac{7}{{\sqrt 2 }}} \right)$$
Work Step by Step
$$\eqalign{
& \left( {7,\frac{\pi }{4},\frac{{3\pi }}{4}} \right) \cr
& {\text{spherical}}\left( {\rho ,\theta ,\phi } \right):\left( {7,\frac{\pi }{4},\frac{{3\pi }}{4}} \right) \to \rho = 7,{\text{ }}\theta = \frac{\pi }{4},{\text{ }}\phi = \frac{{3\pi }}{4} \cr
& {\text{Spherical to cylindrical }}\left( {r,\theta ,z} \right),{\text{ }}\left( {r \geqslant 0} \right){\text{ See page 807}} \cr
& {r^2} = {\rho ^2}{\sin ^2}\phi ,{\text{ }}\theta = \theta ,{\text{ }}z = \rho \cos \phi \cr
& {r^2} = {\left( 7 \right)^2}{\sin ^2}\left( {\frac{{3\pi }}{4}} \right) = \frac{{49}}{2} \to r = \frac{7}{{\sqrt 2 }} \cr
& \theta = \frac{\pi }{4} \cr
& z = 7cos\left( {\frac{{3\pi }}{4}} \right) = - \frac{7}{{\sqrt 2 }} \cr
& {\text{The cylindrical }}\left( {r,\theta ,z} \right){\text{ coordinates are:}} \cr
& \left( {\frac{7}{{\sqrt 2 }},\frac{\pi }{4}, - \frac{7}{{\sqrt 2 }}} \right) \cr} $$
