If the maximum value of a, for which the function ${\mathrm{f}}_{\mathrm{a}}\left(\mathrm{x}\right)={\tan }^{-1}2\mathrm{x}-3\mathrm{ax}+7$ is non-decreasing in $\left(-\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{6}\right),\text{ is }\overline{\mathrm{a}},\text{ then }{\mathrm{f}}_{\overline{\mathrm{a}}}\left(\frac{\mathrm{\pi }}{8}\right)$ is equal to

$\begin{array}{l}{\mathrm{f}}_{\mathrm{a}}\left(\mathrm{x}\right)={\tan }^{-1}2\mathrm{x}-3\mathrm{ax}+7\\ {\mathrm{f}}_{\mathrm{a}}^{\text{'}}\left(\mathrm{x}\right)=\frac{2}{1+4{\mathrm{x}}^2}-3\mathrm{a}\ge 0\\ \mathrm{a}\le {\left(\frac{2}{3\left(1+4{\mathrm{x}}^2\right)}\right)}_{min.}\text{ at }\mathrm{x}=\pm \frac{\mathrm{\pi }}{6}\\ {\mathrm{a}}_{max}=\overline{\mathrm{a}}=\frac{6}{9+{\mathrm{\pi }}^2}\\ {\mathrm{f}}_{\overline{\mathrm{a}}}\left(\frac{\mathrm{\pi }}{8}\right)={\tan }^{-1}\frac{\mathrm{\pi }}{4}-3\frac{6}{9+{\mathrm{\pi }}^2}\frac{\mathrm{\pi }}{8}+7={\tan }^{-1}\frac{\mathrm{\pi }}{4}-\frac{9\mathrm{\pi }}{4\left({\mathrm{\pi }}^2+9\right)}+7\end{array}$

Note:  Answer given by NTA is  $8-\frac{9\mathrm{\pi }}{4\left({\mathrm{\pi }}^2+9\right)}$