Cho ba số thực $x,y,z$ không âm thỏa mãn...

${{2}^{x}}+{{4}^{y}}+{{8}^{z}}=4\Leftrightarrow {{2}^{x}}+{{2}^{2y}}+{{2}^{3z}}=4$.
Ta có $4={{2}^{x}}+{{2}^{2y}}+{{2}^{3z}}\ge 3.\sqrt[3]{{{2}^{x+2y+3z}}}\Leftrightarrow {{2}^{x+2y+3z}}\le \dfrac{64}{27}\Leftrightarrow x+2y+3z\le 6-3{{\log }_{2}}3$.
Khi đó $S\le 1-\dfrac{1}{2}{{\log }_{2}}3$. Suy ra $M=1-\dfrac{1}{2}{{\log }_{2}}3$.
Đặt $\left\{ \begin{aligned}
& a={{2}^{x}}\Rightarrow x={{\log }_{2}}a \\
& b={{4}^{y}}={{2}^{2y}}\Rightarrow y=\dfrac{1}{2}{{\log }_{2}}b \\
& c={{8}^{y}}={{2}^{3z}}\Rightarrow z=\dfrac{1}{3}{{\log }_{2}}b \\
\end{aligned} \right. $, khi đó $ a+b+c=4\Rightarrow b+c=4-a$.
Do $x,y,z\ge 0$ nên $a,b,c\ge 1$, ta có
$\left( b-1 \right)\left( c-1 \right)\ge 0\Leftrightarrow bc\ge \left( b+c \right)-1\Leftrightarrow bc\ge 4-a-1\Leftrightarrow bc\ge 3-a\Leftrightarrow abc\ge a\left( 3-a \right)$.
Xét $f\left( x \right)=3a-{{a}^{2}}$ đạt GTNN trên $a\in \left[ 1;+\infty \right)$ là $f\left( \dfrac{3}{2} \right)=\dfrac{9}{4}$.
Suy ra $abc\ge 3a-{{a}^{2}}\ge \dfrac{9}{4}$.
Mặt khác $S=\dfrac{x}{6}+\dfrac{y}{3}+\dfrac{z}{2}=\dfrac{1}{6}{{\log }_{2}}\left( abc \right)\ge \dfrac{1}{6}{{\log }_{2}}\left( \dfrac{9}{4} \right)$.
Khi đó $N=\dfrac{1}{6}\left( 2{{\log }_{2}}3-2 \right)=\dfrac{1}{3}{{\log }_{2}}3-\dfrac{1}{3}$.
Vậy $T=2M+6N=2\left( 1-\dfrac{1}{2}{{\log }_{2}}3 \right)+6\left( \dfrac{1}{3}{{\log }_{2}}3-\dfrac{1}{3} \right)={{\log }_{2}}3\xrightarrow{{}}T\in \left( 1;2 \right)$.