Answer
$0$
Work Step by Step
The flux through a surface can be defined only when the surface is orientable.
We know that $\iint_S F \cdot dS=\iint_S F \cdot n dS$
Here, $n$ denotes the unit vector.
Since, $\iint_S F \cdot dS=\iint_D [P \dfrac{\partial h}{\partial x}-Q+R\dfrac{\partial h}{\partial z}]dA $
where, $D$ is projection of $S$ onto xz-plane.
$\iint_S F \cdot n dS=\iint_D 125 \sin^2 \phi [5 \cos^2 \theta \cos \phi \sin \phi +\sin \theta \cos \theta \sin \phi +\cos \phi \sin \theta] dA$
$F_1= \int_{0}^{\pi} \int_0^{\pi} 125 \sin^2 \phi [5 \cos^2 \theta \cos \phi \sin \phi +\sin \theta \cos \theta \sin \phi +\cos \phi \sin \theta] dA$
and
$F_2= \int_{0}^{\pi} \int_0^{\pi} 125 \sin^2 \phi [5 \cos^2 \theta (- \cos \phi) \sin \phi +\sin \theta (-\cos \theta) \sin \phi + (-\cos \phi) \times \sin \theta] d\theta d\phi$
Hence, $Flux=F_1+F_2=0$
