The equation of line of shortest  distance between the lines x−31=y−5−2=z−71;x+17=y+1−6=z+11   is

Let  P(α+3, −2α+5, α+7) and Q(7β−1, −6β−1, β−1) be the points on the given lines so that PQ is the line of shortest distance between the given linesD.r’s of PQ=(α−7β+4, −2α+6β+6, α−β+8)Since PQ is perpendicular to the given lines 1(α−7β+4)−2(−2α+6β+6)+1(α−β+8)=0⇒6α−20β=0⇒3α−10β=0 --- (1)and,⇒7(α−7β+4)−6(−2α+6β+6)+1(α−β+8)=0⇒7α−49β+28+12α−36β−36+α−β+8=0 ⇒20α−86β=0⇒10α=43β--- (2)From (1) and (2) α=0, β=0P=(3, 5, 7), Q=(−1, −1, −1)D.r’s of PQ = (–4, –6, –8) = (2, 3, 4)Equation of PQ is x−32=y−53=z−74