Dipole moment of symmetrical molecules
Yes, the statement in your book is very vague.  Molecules are often classified by the symmetry elements they contain.  For example, $\ce{BH3}$ contains a $C_3$ axis perpendicular to the plane containing the $\ce{BH3}$ molecule.  "$C_3$" means that if you rotate the molecule around this axis by $360^\circ/3 = 120^\circ$, you'll get a molecule that is indistinguishable from the molecule you started with.  $\ce{BH3}$ also contains 3 $\sigma_\mathrm{v}$ planes of symmetry.  These planes contain the $C_3$ axis and one of the $\ce{B-H}$ bonds.  $\ce{BH3}$ also contains one $\sigma_\mathrm{h}$ symmetry plane that is perpendicular to the $C_3$ axis and contains the $\ce{BH3}$ molecule.  In addition to these symmetry elements, a number of other symmetry elements exist as well.  

It turns out that only molecules that belong to symmetry classes (point groups) 


$C_n$ (the molecule only contains a $C_n$ axis
$C_{n\mathrm v}$ (the molecule contains a $C_n$ axis and $n$ total $\sigma_\mathrm{v}$ planes)
$C_{\mathrm s}$ (the molecule only has a plane of symmetry that contains
the entire molecule)


can have a permanent dipole moment.  If the molecule belongs to any other point group it cannot have a dipole moment.  Since $\ce{BH3}$ contains a $\sigma_\mathrm{h}$ plane of symmetry ($\ce{BH3}$ belongs to point group $D_{3\mathrm h}$) - it therefore does not belong to groups $C_n$, $C_{n\mathrm v}$ or $C_{\mathrm s}$ - and cannot have a dipole moment.