If the equation $x^2 {\tan }^2\theta – 2\tan \theta .x + 1 = 0$ and  $\left(\frac{1}{1+{\log }_{b}ac}\right)x^2+\left(\frac{1}{1+{\log }_{c}ab}\right)x+(\frac{1}{1+{\log }_{a}bc}-1)=0$ (where a, b, c > 1) have a common root and the $2^{nd}$ equation has equal roots, then the number of possible values of ‘$\theta$’ in $[0, 3\pi ]$ is

$$\begin{gathered} x^2 {\tan }^2\theta – 2\tan \theta .x + 1 = 0\to \left(1\right) \\ \Rightarrow {\left(x\tan \theta -1\right)}^2=0 \\ \Rightarrow x=cot\theta ,cot\theta \\ \left(\frac{1}{1+{\log }_{b}ac}\right)x^2+\left(\frac{1}{1+{\log }_{c}ab}\right)x+(\frac{1}{1+{\log }_{a}bc}-1)=0\to \left(2\right) \\ \Rightarrow its roots are 1,1 \\ \sin ce \left(\frac{1}{1+{\log }_{b}ac}\right)+\left(\frac{1}{1+{\log }_{c}ab}\right)+\left(\frac{1}{1+{\log }_{a}bc}\right)=1 \\ \left(1\right), \left(2\right) root have a common root and \left(2\right) have equal roots \\ \Rightarrow \tan \theta =1\Rightarrow \theta =\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4} \end{gathered}$$