The value of $$\sum_{r=0}^{6} {^{6} C_{r}} \cdot {^{6} C_{6-r}}$$ is equal to:

We need to find $$\sum_{r=0}^{6} \binom{6}{r} \cdot \binom{6}{6-r}$$.

Since $$\binom{6}{6-r} = \binom{6}{r}$$ (by the symmetry property of binomial coefficients), we can rewrite this as $$\sum_{r=0}^{6} \binom{6}{r} \cdot \binom{6}{r} = \sum_{r=0}^{6} \binom{6}{r}^2$$.

Now we use the Vandermonde identity. The identity states that $$\sum_{r=0}^{n} \binom{n}{r} \cdot \binom{n}{n-r} = \binom{2n}{n}$$. This comes from comparing the coefficient of $$x^n$$ on both sides of $$(1+x)^n \cdot (1+x)^n = (1+x)^{2n}$$.

Here $$n = 6$$, so $$\sum_{r=0}^{6} \binom{6}{r} \cdot \binom{6}{6-r} = \binom{12}{6}$$.

Computing $$\binom{12}{6} = \frac{12!}{6! \cdot 6!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{665280}{720} = 924$$.

The answer is $$924$$, which is Option D.