Ta có: $\begin{array}{*{20}{l}} {\left\{ {\begin{array}{*{20}{l}} {x = 2017 - t}\\ {y = 1 + 3t} \end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}} {3x = 6051 - 3t\left( 1 \right)}\\ { - y = - 1 - 3t\left( 2 \right)} \end{array}} \right.}\\ {\left( 1 \right) + \left( 2 \right) \Rightarrow 3x + y = 6052} \end{array}$
Ta có hệ phương trình: $\left\{ {\begin{array}{*{20}{l}} {x - 3y = - 2017}\\ {3x + y = 6052} \end{array}} \right.$
Mà $a_1.a_2+b_1.b_2=0 \Rightarrow 1.3+(-3).1=0$
Vậy ${d_1} \bot {d_2}$
Chọn $C$
