: Đường thẳng D qua ${{M}_{0}}\left( -2;1;-2 \right)$ và có VTCP

         $\overrightarrow{a}=\left( 1;2;-2 \right)$ Þ $\overrightarrow{{{M}_{0}}{{M}_{1}}}=\left( 4;2;2 \right)$

Ta có: $\left[ \overrightarrow{{{M}_{0}}{{M}_{1}}};\overrightarrow{a} \right]=\left( -8;10;6 \right)$

         $\Rightarrow d\left( {{M}_{1}};\Delta  \right)=\frac{\left| \text{ }\,\left[ \overrightarrow{a};\overrightarrow{{{M}_{0}}{{M}_{1}}} \right]\text{ } \right|}{\left| \text{ }\overrightarrow{a}\text{ } \right|}$

$\text{               }=\frac{\sqrt{{{(-8)}^{2}}+{{10}^{2}}+{{6}^{2}}}}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{(-2)}^{2}}}}=\frac{10\sqrt{2}}{3}$

Chọn đáp án A