Trong không gian Oxyz, cho ba đường thẳng d:x/1=y/1=(z+1)/-2

Đáp án A

\(H \in {\Delta _1} \Rightarrow H\left( {2t + 3;t;1 + t} \right);K \in {\Delta _2} \Rightarrow K\left( {1 + s;2 + 2s;s} \right) \Rightarrow \overrightarrow {HK} = \left( {s - 2t - 2;2s - t;s - t - 1} \right)\)

\(\Delta \bot d \Rightarrow \overrightarrow {HK} \bot \overrightarrow {{u_d}} \Rightarrow s - t + 2 = 0 \Leftrightarrow s = t - 2 \Rightarrow H{K^2} = 2{\left( {t + 1} \right)^2} + 27 \ge 27\)

\(t = - 1 \Rightarrow \overrightarrow {HK} = \left( { - 3; - 3; - 3} \right) = - 3\left( {1;1;1} \right) \Rightarrow h - k = 0\)