The correct statement(s) about intermolecular forces is(are)

Intermolecular interaction energies are obtained by applying Coulomb’s law to the various charge distributions that can exist on (or be induced in) molecules - permanent charges, permanent dipoles, induced dipoles, etc. The distance dependence and, in some cases, the temperature dependence of the average energy decide whether a given statement is true or false.

Option A
• Potential energy between two point charges (say $$q_1$$ and $$q_2$$) is $$U_{qq}= \dfrac{q_1q_2}{4\pi\varepsilon_0\,r} \propto \dfrac{1}{r}$$.
• Potential energy between a point charge $$q$$ and a point dipole of moment $$\mu$$ (the dipole oriented so that the interaction is maximum) is $$U_{q\mu}= -\dfrac{q\mu}{4\pi\varepsilon_0\,r^{2}} \propto \dfrac{1}{r^{2}}$$.
Because $$1/r^{2}$$ tends to zero faster than $$1/r$$ when $$r\rightarrow\infty$$, the charge-dipole energy approaches zero more rapidly than the charge-charge energy. Option A claims the opposite, hence it is false.

Option B
For two freely rotating permanent dipoles, the orientation-averaged (thermal average) interaction energy has Keesom form
$$\langle U_{\mu\mu}\rangle = -\dfrac{2\mu_1^{2}\mu_2^{2}}{3(4\pi\varepsilon_0)^{2}k_{\mathrm B}T\,r^{6}} \propto -\dfrac{1}{T\,r^{6}}.$$
Thus the energy falls off as $$1/r^{6}$$, not $$1/r^{3}$$. Therefore Option B is false.

Option C
Dipole-induced dipole interaction (Debye interaction): a permanent dipole $$\mu$$ induces a dipole $$\alpha \,E$$ in a neighbouring polarisable molecule (where $$\alpha$$ is its polarisability and $$E$$ the electric field of the permanent dipole). The orientation-averaged energy is
$$\langle U_{\mu\text{-ind}}\rangle = -\dfrac{\mu^{2}\alpha}{(4\pi\varepsilon_0)^{2}\,2\,r^{6}},$$
which contains no temperature term. Hence it is independent of temperature. Option C is true.

Option D
Even molecules with no permanent dipole can attract each other through London dispersion forces. A momentary (instantaneous) dipole in one molecule induces a dipole in the other, giving an average interaction energy
$$\langle U_{\text{disp}}\rangle = -\dfrac{3h\nu\,\alpha_1\alpha_2}{4(4\pi\varepsilon_0)^{2}\,r^{6}},$$
which is always attractive. Hence non-polar molecules do attract one another despite lacking permanent dipoles. Option D is true.

Therefore the correct statements are:
Option C (dipole-induced dipole energy is temperature-independent) and
Option D (non-polar molecules attract each other via dispersion forces).

Final answer: Option C and Option D.