To determine which option is correct based on the principle of homogeneity of dimensions, we need to ensure that both sides of the equation have the same dimensions. The time period ($$T$$) is measured in units of time ($$T$$), the gravitational constant ($$G$$) has units of $$\text{m}^3\text{kg}^{-1}\text{s}^{-2}$$, mass ($$M$$) has units of mass ($$M$$), and the radius of orbit ($$r$$) has units of length ($$L$$).

Let's analyze each option:

Option A: $$T^2=\frac{4 \pi^2 r}{G M^2}$$

The dimensions of the right-hand side of the equation are $$\frac{L}{\left(\frac{L^3}{MT^2}\right)M^2}=\frac{L}{L^3M^{-1}T^{-2}M^2}=\frac{L}{L^3T^{-2}}=L^{-2}T^2$$, which do not match with $$T^2$$ (time squared). Thus, Option A is incorrect.

Option B: $$T^2=4 \pi^2 r^3$$

The dimensions of the right-hand side are $$L^3$$, which clearly do not match the dimensions $$T^2$$ of the squared time period. So, Option B is incorrect.

Option C: $$T^2=\frac{4 \pi^2 r^3}{G M}$$

The dimensions of the right-hand side of the equation are $$\frac{L^3}{\left(\frac{L^3}{MT^2}\right)M}=\frac{L^3}{L^3T^{-2}}=T^2$$, which match the dimensions of the squared time period $$T^2$$. Therefore, Option C is correct based on the principle of homogeneity of dimensions.

Option D: $$T^2=\frac{4 \pi^2 r^2}{G M}$$

The dimensions of the right-hand side are $$\frac{L^2}{\left(\frac{L^3}{MT^2}\right)M}=\frac{L^2}{L^3M^{-1}T^{-2}M}=L^{-1}T^2$$, which do not match with $$T^2$$ (time squared). Hence, Option D is incorrect.

Thus, based on the principle of homogeneity of dimensions, Option C is the correct one: $$T^2=\frac{4 \pi^2 r^3}{G M}$$. This equation also corresponds to Kepler's third law of planetary motion, which relates the orbital period of a planet to its orbital radius, considering the mass of the central body.