Still in flux

I will use a combination of symmetry considerations and "brute force." I hope that somebody will come up with a more elegant solution.

The given plane $2x+y-2z=3$ has a distance of $\frac{3}{\sqrt{2^2+1^2+(-2)^2}}=1$ from the origin; we can rotate the plane into $z=1$ without changing the flux. Thus we can consider a hexagonal surface with its vertices at $z=1$ and $x^2+y^2=1$ . We can place two vertices at $P(\cos\alpha,\sin\alpha,1)$ and $Q(\cos\alpha,-\sin\alpha,1)$ where $\alpha=\frac{\pi}{6}$ . The flux $\Phi$ we seek is six times the upward flux through the equilateral triangle with its vertices at $P,Q$ and $(0,0,1)$ .

The upward flux density through the plane $z=1$ is $\vec{F} \cdot \vec{n}=\vec{F} \cdot \vec{k}=\frac{1}{(x^2+y^2+1)^{3/2}}$ . Thus $\Phi=6\int_{0}^{\cos \alpha}\int_{-(\tan \alpha)x}^{(\tan \alpha)x}\frac{1}{(x^2+y^2+1)^{3/2}}dy\ dx=12\arctan(\sqrt{6})-4\pi\approx 1.63203$

The required answer is $\boxed{1632}$ .