If $a, b, c$ are in $\mathrm{GP}$ and $\log a-\log 2b, \log 2b-\log 3c$ and $\log 3c-\log a$ are in $\mathrm{AP}$, then $a, b$ and $c$ are the lengths of the sides of a triangle, which is

Given, $a, b, c$ are in $\mathrm{GP}$.

$\therefore b^2=ac$

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

$\Rightarrow b^2=ac$ and $2b=3c$

$\because a+b=\frac{5a}{3}>c, b+c=\frac{10a}{9}>a$

and $c+a=\frac{13a}{9}>b$

$\therefore a, b, c$ are the sides of a triangle.

Also, $a$ is the greatest side

$\therefore \cos A=\frac{b^2+c^2-a^2}{2bc}=-\frac{29}{48}<0$

$\therefore ∆\mathrm{ABC}$ is an obtuse angled triangle.

Hence, option (d) is correct.