Trong không gian Oxyz, cho hai đường thẳng d1 : x=2 y=-1+tz=-2=2t

Đáp án B

Gọi $A\left(2;-1+a;-2+2a\right)=\Delta \cap d_1;B\left(1-b;-1+b;3+b\right)=\Delta \cap d_2$.

Suy ra $\overrightarrow{AB}=\left(-b-1;b-a;b-2a+5\right)$. Theo giả thiết ta có

$\iff \left\{\begin{array}{l}\overrightarrow{u_{d_1}}.\overrightarrow{AB}=0\\ \overrightarrow{u_{d_2}}.\overrightarrow{AB}=0\end{array}\right.\iff \left\{\begin{array}{l}b-a+2\left(b-2a+5\right)=0\\ b+1+b-a+b-2a+5=0\end{array}\right.\iff \left\{\begin{array}{l}a=2\\ b=0\end{array}\right.\Rightarrow \left\{\begin{array}{l}B\left(1;-1;3\right)\\ \overrightarrow{AB}=\left(-1;-2;1\right)\end{array}\right.$.

Vậy phương trình $\Delta :\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-3}{-1}$.