Explanation

(b)
Since, $a,\ b,\ c$ are in GP.
$\Rightarrow \ b^{2} = ac$
And $\log{a - \log{2b,\log{2b - \log{3c}\ and\log{3c - \log a\ }}}}$are in AP.
$\Rightarrow 2(\log{2b - \log{3c) = \log{a - \log{2b + \log{3c - \log a}}}}}$
$\therefore\ b^{2} = ac\ and\ 2b = 3c$
$\Rightarrow \ b = \frac{2a}{3}\ \ and\ c = \frac{4a}{9}$
Since, $a + b = \frac{5a}{3} > c,\ b + c = \frac{10a}{9} > a,$
$c + a = \frac{13a}{9} > b$
It implies that $a,\ b,\ c$ form a triangle with $a$ as the greatest side.
Now, let us find the greatest angle $A\ of\ \mathrm{\Delta}\ ABC$ by using the cosine formula.
$\cos{A = \frac{b^{2} + c^{2} - a^{2}}{2bc} = \frac{\frac{4a^{2}}{9} + \frac{16a^{2}}{81} - a^{2}}{\frac{4a}{3}.\frac{4a}{9}}}$
$= \ - \frac{29}{48} < 0$
$\therefore\ \$The angle $A$ is obtuse.