\(\log \mathrm{a}, \log \mathrm{b}, \log \mathrm{c}\) are in A.P.

\(\begin{aligned}

&\Rightarrow 2 \log \mathrm{b}=\log \mathrm{a}+\log \mathrm{c} \\

&\Rightarrow \log \mathrm{b}^{2}=\log (\mathrm{ac}) \\

&\Rightarrow \mathrm{b}^{2}=\mathrm{ac} \Rightarrow \mathrm{a}, \mathrm{b}, \mathrm{c} \text { are in G.P. } \\

&\log \mathrm{a}-\log 2 \mathrm{~b}, \log 2 \mathrm{~b}-\log 3 \mathrm{c}, \log 3 \mathrm{c}-\log \text { a are in A.P. }

\end{aligned}\)

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

\(\Rightarrow 3 \log 2 \mathrm{~b}=3 \log 3 \mathrm{c} \Rightarrow 2 \mathrm{~b}=3 \mathrm{c}\)

Now, \(b^{2}=a c \Rightarrow b^{2}=a \cdot \frac{2 b}{3} \Rightarrow b=\frac{2 a}{3}, c=\frac{4 a}{9}\)

\(\begin{aligned}

&\text { i.e., } a=a, b=\frac{2 a}{3}, c=\frac{4 a}{9} \\

&\Rightarrow a: b: c=1: \frac{2}{3}: \frac{4}{9}=9: 6: 4

\end{aligned}\)

Since, sum of any two is greater than the \(3^{\mathrm{rd}}, \mathrm{a}, \mathrm{b}, \mathrm{c}\) form a triangle.