1. Since $\log_e a,\log_e b,\log_e c$ are in A.P., the middle term is the average of the other two:

$2\log b = \log a + \log c$

Using log properties,

$\log b^2 = \log(ac) \implies b^2 = ac \quad ...(1)$

2. Now consider the second A.P.:

$\log a - \log 2b,\quad \log 2b - \log 3c,\quad \log 3c - \log a$

First simplify each term:

$\log a - \log 2b = \log\left(\frac{a}{2b}\right)$ $\log 2b - \log 3c = \log\left(\frac{2b}{3c}\right)$ $\log 3c - \log a = \log\left(\frac{3c}{a}\right)$

Since these are in A.P.,

$2\log\left(\frac{2b}{3c}\right)=\log\left(\frac{a}{2b}\right)+\log\left(\frac{3c}{a}\right)$

Combine logs:

$\log\left(\frac{4b^2}{9c^2}\right)=\log\left(\frac{a}{2b}\cdot \frac{3c}{a}\right)$

$\log\left(\frac{4b^2}{9c^2}\right)=\log\left(\frac{3c}{2b}\right)$

Hence,

$\frac{4b^2}{9c^2}=\frac{3c}{2b}$

Cross-multiplying,

$8b^3 = 27c^3$

$\left(\frac{b}{c}\right)^3 = \left(\frac{3}{2}\right)^3$

So,

$\frac{b}{c} = \frac{3}{2} \quad ...(2)$

3. Using $(1)$, $b^2 = ac$, so

$a = \frac{b^2}{c}$

From $b:c = 3:2$, let

$b=3k,\quad c=2k$

Then

$a = \frac{(3k)^2}{2k} = \frac{9k}{2}$

Thus,

$a:b:c = \frac{9k}{2}:3k:2k$

Multiply by $2$:

$a:b:c = 9:6:4$

4. Therefore the correct option is:

$\boxed{B\; (9:6:4)}$