b
Let angle between $\overrightarrow{\mathrm{a}} \& \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{b}} \& \overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{c}} \& \overrightarrow{\mathrm{a}}$ are $\theta_{1}, \theta_{2}, \theta_{3}$ respectively.
$\because$ length of projection are in $G.P.$
$\therefore \quad {\left( {\frac{{\vec b \cdot \vec c}}{{|{\rm{c}}|}}} \right)^2} = \left( {\frac{{\vec a \cdot \vec b}}{{|{\rm{b}}|}}} \right)\left( {\frac{{\vec c \cdot \vec a}}{{|{\rm{a}}|}}} \right)$
$\Rightarrow \quad \frac{|b|^{2}|c|^{2} \cos ^{2} \theta_{2}}{|c|^{2}}=\frac{|a||b| \cos \theta_{1}|c||a|}{|b|} \cos \theta_{3}$
$\Rightarrow \quad \cos ^{2} \theta_{2}=\frac{|a||c|}{|b|^{2}} \times \cos \theta_{1} \cos \theta_{3}$
$\Rightarrow \quad \cos ^{2} \theta_{2}=\frac{2 \times 9}{9} \times \frac{1}{2}\left[\cos \left(\theta_{1}+\theta_{3}\right)+\cos \left(\theta_{1}-\theta_{3}\right)\right]$
$\left(\theta_{1}=\frac{5 \pi}{12} ; \theta_{3}=\frac{\pi}{12} \Rightarrow \theta_{1}+\theta_{3}=\frac{\pi}{2} ; \theta_{1}-\theta_{3}=\frac{\pi}{3}\right)$
$\Rightarrow \quad \cos ^{2} \theta_{2}=\frac{1}{2} \Rightarrow \theta_{2}=\frac{\pi}{4}$