If $$\log_e a, \log_e b, \log_e c$$ are in an A.P. and $$\log_e a - \log_e 2b, \log_e 2b - \log_e 3c, \log_e 3c - \log_e a$$ are also in an A.P., then $$a : b : c$$ is equal to

We are given that $$\log_e a, \log_e b, \log_e c$$ are in A.P. and $$\log_e a - \log_e 2b, \log_e 2b - \log_e 3c, \log_e 3c - \log_e a$$ are also in A.P. We need to find $$a : b : c$$.

Since $$\log_e a, \log_e b, \log_e c$$ are in A.P., we have $$2\log_e b = \log_e a + \log_e c$$, which implies $$\log_e b^2 = \log_e(ac)$$ and hence $$b^2 = ac \quad \cdots (1)$$.

Similarly, because $$\log_e a - \log_e 2b$$, $$\log_e 2b - \log_e 3c$$, and $$\log_e 3c - \log_e a$$ are in A.P., it follows that $$2(\log_e 2b - \log_e 3c) = (\log_e a - \log_e 2b) + (\log_e 3c - \log_e a)$$. Simplifying the right side gives $$(\log_e a - \log_e 2b) + (\log_e 3c - \log_e a) = \log_e 3c - \log_e 2b = \log_e\frac{3c}{2b}$$, while the left side is $$2\log_e\frac{2b}{3c} = \log_e\left(\frac{2b}{3c}\right)^2$$. Equating these yields $$\log_e\left(\frac{2b}{3c}\right)^2 = \log_e\frac{3c}{2b}$$, so $$\left(\frac{2b}{3c}\right)^2 = \frac{3c}{2b}$$ and hence $$\left(\frac{2b}{3c}\right)^3 = 1$$, giving $$\frac{2b}{3c} = 1$$ or $$2b = 3c$$, i.e. $$c = \frac{2b}{3} \quad \cdots (2)$$.

Substituting $$c = \frac{2b}{3}$$ from (2) into (1) gives $$b^2 = a \cdot \frac{2b}{3}$$, so $$a = \frac{3b}{2} \quad \cdots (3)$$.

Therefore, $$a : b : c = \frac{3b}{2} : b : \frac{2b}{3}$$, and multiplying through by 6 yields $$a : b : c = 9 : 6 : 4$$. Hence the required ratio is 9 : 6 : 4.

Therefore, the correct answer is Option 1: 9 : 6 : 4.