Chọn : Đáp án A

Phương trình tham số của hai đường thẳng :

$\left( {{d}_{1}} \right)=\left\{ \begin{align}  & x={{t}_{1}} \\  & y=1-{{t}_{1}} \\  & z=-2{{t}_{1}} \\ \end{align} \right.\Rightarrow \overrightarrow{{{u}_{1}}}=\left( 1;-1;-2 \right);{{M}_{1}}\left( 0;1;0 \right)\in {{d}_{1}}$

$\left( {{d}_{2}} \right)=\left\{ \begin{align}  & x={{t}_{2}} \\  & y=1-2{{t}_{2}} \\  & z=2 \\ \end{align} \right.\Rightarrow \overrightarrow{{{u}_{2}}}=\left( 1;-2;0 \right);{{M}_{2}}\left( 0;1;2 \right)\in {{d}_{2}}$

$\Rightarrow \overrightarrow{{{M}_{1}}{{M}_{2}}}=\left( 0;0;2 \right)$

Vecto chỉ phương của đường vuông góc chung : $\overrightarrow{u}=\left[ \overrightarrow{{{u}_{2}}};\overrightarrow{{{u}_{1}}} \right]=\left( 4;2;1 \right)$

$\Rightarrow \overrightarrow{u}.\overrightarrow{{{M}_{1}}{{M}_{2}}}=2\ne 0\Rightarrow {{d}_{1}};{{d}_{2}}$ chéo nhau.

Gọi $\left( \Delta  \right)$ cắt ${{d}_{1}}$ tại $M\Rightarrow M\left( {{t}_{1}};1-{{t}_{1}};-2{{t}_{1}} \right);\left( \Delta  \right)$ cắt ${{d}_{2}}$ tại $N\Rightarrow N\left( {{t}_{2}};1-2{{t}_{2}};2 \right)$

$\Rightarrow \overrightarrow{MN}=\left( {{t}_{2}}-{{t}_{1}};{{t}_{1}}-2{{t}_{2}};2+2{{t}_{1}} \right)$

$\overrightarrow{MN}$//$\overrightarrow{u}$$\Rightarrow \frac{{{t}_{2}}-{{t}_{1}}}{4}=\frac{{{t}_{1}}-2{{t}_{2}}}{2}=\frac{2+2{{t}_{1}}}{1}\Rightarrow \left\{ \begin{align}  & 6{{t}_{1}}-10{{t}_{2}}=0 \\  & 3{{t}_{1}}+2{{t}_{2}}=-4 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}  & {{t}_{1}}=-\frac{20}{21} \\  & {{t}_{2}}=-\frac{4}{7} \\ \end{align} \right.\Rightarrow N\left( -\frac{4}{7};\frac{15}{7};2 \right)$

$\Rightarrow \left( \Delta  \right)\left\{ \begin{align}  & x=-\frac{4}{7}+4t \\  & y=\frac{15}{7}+2t\left( t\in \mathbb{R} \right) \\  & z=2+t \\ \end{align} \right.$