Answer
$(2,\frac{\pi}{3},\frac{\pi}{3})$.
Work Step by Step
Since $ x=\frac{\sqrt3}{2}, \, y=\frac{3}{2}, \, z=1$ and
$$\rho=\sqrt{x^2+y^2+z^2},\quad \theta=\tan^{-1}( y/x), \quad \phi=\cos^{-1}(z/\rho),$$
then we have
$$\rho=\sqrt{\frac{3}{4}+\frac{9}{4}+1^2}= 2, $$
$$\theta=\tan^{-1}( 3/\sqrt 3)=\frac{\pi}{3}, $$
$$\phi=\cos^{-1}(1/2)=\frac{\pi}{3}.$$
Hence, the spherical coordinates are $(2,\frac{\pi}{3},\frac{\pi}{3})$.
