Why is the vanadium(3+) ion paramagnetic?
In addition to the general rules of how electronic configurations of atoms and ions are calculated, the elements from the $\mathrm{d}$-block (a.k.a. the transition metals) obey one special rule:

In general, electrons are removed from the valence-shell $\mathrm{s}$-orbitals before they are removed from valence $\mathrm{d}$-orbitals when transition metals are ionized.

(I took this formulation from these online lecture notes, but you will find equivalent statements in your textbooks.)
So, what that does mean is that if you remove electrons from vanadium(0), you will remove the $\mathrm{4s}$ electrons before you remove the $\mathrm{3d}$-electrons. So, you have the following electronic configurations:

$\ce{V}$ is $\ce{[Ar]} \mathrm{4s^2 3d^3}$
$\ce{V^2+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^3}$
$\ce{V^3+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^2}$
$\ce{V^4+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^1}$
$\ce{V^5+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^0}$

And thus, $\ce{V^3+}$ is paramagnetic, because it has two unpaired $\mathrm{3d}$-electrons. In fact, all the ions above are paramagnetic, except $\ce{V^5+}$.