Phương trình mặt phẳng (P) chứa đường thẳng ${{\Delta }_{1}}$có dạng:

$\begin{align} & \alpha \left( x-2y+z-4 \right)+\beta \left( x+2y-2z+4 \right)=0\left( {{\alpha }^{2}}+{{\beta }^{2}}\ne 0 \right) \\ & a\left( x-2y+z-4 \right)+b\left( x+2y-2z+4 \right)=0\left( {{a}^{2}}+{{b}^{2}}\ne 0 \right) \\ & \Leftrightarrow \left( a+b \right)x-\left( 2a-2b \right)y+\left( a-2b \right)z-4a+4b=0 \\ \end{align}$

Vậy  $\overrightarrow{{{n}_{p}}}=\left( a+b;-2a+2b;a-2b \right)$Ta có: $\left\{ \begin{align} & \overrightarrow{{{u}_{2}}}=\left( 1;1;2 \right)||{{\Delta }_{1}} \\ & {{M}_{2}}\left( 1;2;1 \right)\in {{\Delta }_{2}} \\ \end{align} \right.$

$(P)||{{\Delta }_{2}}\left\{ \begin{align} & \overrightarrow{{{n}_{p}}}.\overrightarrow{{{u}_{2}}}=0 \\ & {{M}_{2}}\left( 1;2;1 \right)\notin (P) \\ \end{align} \right.$ $\Leftrightarrow \left\{ \begin{align} & a-b=0 \\ & {{M}_{2}}\notin (P) \\ \end{align} \right.$

Vậy (P): $2x-z=0$

Vậy đáp án đúng là A