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

$\text {Given, }$ <br/><br/>$$ \begin{aligned} f_a(x) & =\tan ^{-1} 2 x-3 a x+7 \\\\ f_a^{\prime}(x) & =\frac{2}{1+4 x^2}-3 a \end{aligned} $$ <br/><br/>As the function $f_a^{\prime}(x)$ is non-decreasing <br/><br/>$$ \begin{aligned} & \text { in }\left(-\frac{\pi}{6}, \frac{\pi}{6}\right), \\\\ & f_a^{\prime}(x) \geq 0 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \Rightarrow \frac{2}{1+4 x^2}-3 a \geq 0 \Rightarrow \frac{2}{1+4 x^2} \geq 3 a \\\\ & \Rightarrow a \leq \frac{2}{3\left(1+4 x^2\right)}, \text { when } x \in\left(-\frac{\pi}{6}, \frac{\pi}{6}\right) \\\\ & \because a \text { is maximum when } x^2=\frac{\pi^2}{36}, \end{aligned} $$ <br/><br/>$$ \begin{aligned} a_{\max } & =\frac{2}{3\left(1+\frac{4 \pi^2}{36}\right)}=\frac{2 \times 12}{36+4 \pi^2} \\\\ & =\frac{6}{9+\pi^2} \\\\ \therefore \bar{a} & =\frac{6}{9+\pi^2} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \therefore f_a(x)=\tan ^{-1} 2 x-\frac{18}{9+\pi^2} x+7 \\\\ & f_\pi\left(\frac{\pi}{8}\right)=\tan ^{-1} 2\left(\frac{\pi}{8}\right)-\frac{18}{9+\pi^2} \times \frac{\pi}{8}+7 \\\\ & =\tan ^{-1} \frac{\pi}{4}-\frac{9 \pi}{36+4 \pi^2}+7 \\\\ & =8-\frac{9 \pi}{36+4 \pi^2}=8-\frac{9 \pi}{4\left(9+\pi^2\right)} \end{aligned} $$