Let $$a_1, a_2, a_3, \ldots$$ be in an A.P. such that $$\sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1$$, $$a_1 \neq 0$$. If $$\sum_{k=1}^{n} a_k = 0$$, then $$n$$ is:

We are given an A.P. $$a_1, a_2, a_3, \ldots$$ with common difference $$d$$, such that $$\sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5}a_1$$ and $$a_1 \neq 0$$.

The terms $$a_{2k-1}$$ for $$k = 1, 2, \ldots, 12$$ are $$a_1, a_3, a_5, \ldots, a_{23}$$. This is an A.P. with first term $$a_1$$, common difference $$2d$$, and 12 terms.

$$\sum_{k=1}^{12} a_{2k-1} = 12a_1 + 2d \cdot \frac{12 \cdot 11}{2} = 12a_1 + 132d$$

Setting this equal to $$-\frac{72}{5}a_1$$:

$$12a_1 + 132d = -\frac{72}{5}a_1$$

$$132d = -\frac{72}{5}a_1 - 12a_1 = -\frac{72 + 60}{5}a_1 = -\frac{132}{5}a_1$$

$$d = -\frac{a_1}{5}$$

Now, $$\sum_{k=1}^{n} a_k = \frac{n}{2}(2a_1 + (n-1)d) = 0$$.

Since $$a_1 \neq 0$$, we need $$2a_1 + (n-1)d = 0$$ (assuming $$n \neq 0$$).

$$2a_1 + (n-1)\left(-\frac{a_1}{5}\right) = 0$$

$$2 - \frac{n-1}{5} = 0$$

$$\frac{n-1}{5} = 2$$

$$n - 1 = 10$$

$$n = 11$$

Hence, the correct answer is Option A.