The number of values of $$a \in N$$ such that the variance of 3, 7, 12, $$a$$, 43 $$- a$$ is a natural number is:

We need to find the number of values of $$a \in \mathbb{N}$$ such that the variance of 3, 7, 12, $$a$$, $$43 - a$$ is a natural number. The mean is $$\bar{x} = \frac{3 + 7 + 12 + a + (43 - a)}{5} = \frac{65}{5} = 13$$, and hence the variance is $$\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{5} = \frac{(3-13)^2 + (7-13)^2 + (12-13)^2 + (a-13)^2 + (43-a-13)^2}{5} = \frac{100 + 36 + 1 + (a-13)^2 + (30-a)^2}{5} = \frac{137 + (a-13)^2 + (30-a)^2}{5}.$$

Now $$(a-13)^2 + (30-a)^2 = a^2 - 26a + 169 + 900 - 60a + a^2 = 2a^2 - 86a + 1069$$ so $$\sigma^2 = \frac{137 + 2a^2 - 86a + 1069}{5} = \frac{2a^2 - 86a + 1206}{5}.$$

For the variance to be a natural number, we need $$\frac{2a^2 - 86a + 1206}{5} \in \mathbb{N}$$, which requires $$2a^2 - 86a + 1206 \equiv 0 \pmod{5}.$$ Reducing modulo 5, since $$86 = 17 \times 5 + 1$$ and $$1206 = 241 \times 5 + 1$$, we get $$2a^2 - 86a + 1206 \equiv 2a^2 - a + 1 \pmod{5}.$$ Checking each residue class mod 5: for $$a \equiv 0$$: $$0 - 0 + 1 = 1 \not\equiv 0$$; for $$a \equiv 1$$: $$2 - 1 + 1 = 2 \not\equiv 0$$; for $$a \equiv 2$$: $$8 - 2 + 1 = 7 \equiv 2 \not\equiv 0$$; for $$a \equiv 3$$: $$18 - 3 + 1 = 16 \equiv 1 \not\equiv 0$$; for $$a \equiv 4$$: $$32 - 4 + 1 = 29 \equiv 4 \not\equiv 0$$.

For no value of $$a \pmod{5}$$ is the expression divisible by 5. Therefore, the variance is never a natural number for any $$a \in \mathbb{N}$$. The number of such values is $$\textbf{0}$$, so the answer is Option A: $$\textbf{0}$$.