The distance between the liner=2i^−2j^+3k^+λ(i^−j^+4k^) and the plane r.(i^+5j^+k^)=5 is

The given line is r=a+tb Where a=2i−2j+3k, b=i−j+4k and given plane is r.n=p, Where n=i^+5j^+k^,p=5 Since b.n=1−5+4=0 ∴ given line is parallel to the given plane ∴ the distance between the line and the plane is equal to length of the perpendicular from the point a=2i−2j+3k on the line to tae given plane.∴ Reqd. distance =|(2i−2j+3k).(i+5j+k)−51+25+1| =|2−10+3−527|=1033.