Equation of first line, 1x−6=2y−2=2z−2=k(say) ∴x=k+6,y=−2k+2,z=2k+2
Hence, general point on the first line, P≡(k+6,−2k+2,2k+2)
Equation of second line, 3x+4=−2y=−2Z+1=l(say) ∴x=3l−4,y=−2l,z=−2l−1
Hence, general point on the second line, Q≡(3l−4,−2l,−2l−1)
Direction ratios of PQ are 3l−4−k−6,−2l+2k−2,−2l−1−2k−2
i.e. 3l−k−10,−2l+2k−2,−2l−2k−3
Now |PQ| will be the shortest distance
between the two lines if PQ is perpendicular
to both the lines. Hence, 1(3l−k−10)+(−2) (−2l+2k−2)+2(−2l−2k−3)=0
and 3(3l−k−10)+(−2)(−2l+2k−2)+(−2)(−21−2k−3)=0 i.e.3l−9k=12orl−3k=4...(i)
and 17l−3k=20...(ii)
Subtracting equation (i) from(ii), we get 161=16∴l=1
Putting this value of l in equation \left(i\right), we get −3k=3,∴k=−1 ∴P≡(−1+6,−2(−1)+2,2(−1)+2) ≡(5,4,0)
Similarly, Q=(−1,−2,−3)
Hence, shortest distance, PQ,
=(−1−5)2+(−2−4)2+(−3−0)2 =(−6)2+(−6)2+(−3)2=36+36+9
=9 units