Two values of x for which the sixth term of the expansion of $E={\left(3^{{\log }_3\sqrt{9^{|x-2|}}}+7^{(1/5){\log }_7\left[\left(4\right)\cdot 3^{|x-2|}-9\right]}\right)}^7$ is 567, is

Put  $y=3^{{\log }_3\sqrt{9^{|x-2|}}}\Rightarrow {\log }_{3}y={\log }_3\sqrt{9^{|x-2|}}$

$\Rightarrow y=\sqrt{9^{|x-2|}}=3^{|x-2|}$

Next Put $z=\left(7^{(1/5){\log }_7\left[\left(4\right)\cdot 3^{|x-2|}-9\right]}\right)$

$\begin{array}{l}\Rightarrow {\log }_{7}z=\frac{1}{5}{\log }_7\left[\left(4\right)3^{|x-2|}-9\right]={\log }_7{\left[\left(4\right)3^{|x-2|}-9\right]}^{1/5}\\ \Rightarrow z={\left[\left(4\right)3^{|x-2|}-9\right]}^{1/5}\end{array}$

Now, $E=(y+z{)}^7$and the 6th term in its expansion is
given by

$\begin{array}{l}{T}_6{=}^7{C}_5{y}^{7-5}{z}^5=21{\left(3^{|x-2|}\right)}^2\left\{\left(4\right)3^{|x-2|}-9\right\}\\ \Rightarrow 567=21\left\{3^{2x-21}\right\}\left\{\left(4\right)3^{|x-2|}-9\right\}\\ \Rightarrow 27=\left\{\left(4\right)3^{3x-2\mid }-\left(9\right)\left(3^{2|x-2|}\right)\right\}\\ \Rightarrow 27=\left(4\right)3^{3x-2\mid }-\left(9\right)3^{2|x-2|}\end{array}$

$\Rightarrow 4u^3-9u^2-27=0$ where $u=3^{|x-2|}$

Note that u = 3 satisfies this equation*.

$\therefore 3^{|x-2|}=3\Rightarrow |x-2\mid =1\Rightarrow x-2=\pm 1$

$x=2\pm 1=3$ or 1.