If  log a, log b, and log c are in A.P and also $\log a-\log 2b,\log 2b-\log 3c,\log 3c-\log a$  are  in A.P .,then:

$\log a,\log b,\log c$  are in A.P

$\Rightarrow 2\log b=\log a+\log c\Rightarrow \log b^2=\log \left(ac\right)$

$\Rightarrow b^2=ac\Rightarrow a,b,careinG.P.$

$\log a-\log 2b,\log 2b-\log 3c,\log 3c-\log aareinA.P$

$\Rightarrow 2\left(\log 2b-\log 3c\right)=\left(\log a-\log 2b\right)+\left(\log 3c-\log a\right)$

$\Rightarrow 3\log 2b=3\log 3c\Rightarrow 2b=3c$

Now,$b^2=ac\Rightarrow b^2=a.\frac{2b}{3}\Rightarrow b=\frac{2a}{3},c=\frac{4a}{9}$

i.e.  a= a. b$=\frac{2a}{3},c=\frac{4a}{9}\Rightarrow a:b:c=1:\frac{2}{3}:\frac{4}{9}=9:6:4$

since, sum of any two is greater than the 3^rd,a,b,c, form a triangle.