The point on the line  $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$which is nearest to the line $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$is $\left(a,b,c\right)$ where  $a+b+c=$

detailed solution

Correct option is C

The given two lines are not parallel lines, so both possess common perpendicular line. The ends of this common perpendicular are the nearest points to each other The general point on the line x+22=y+63=z−34−10 isP−2+2s,−6+3s,34−10sAnd the general point on the line  x+64=y−7−3=z−7−2isQ−6+4t,7−3t,7−2tThe direction ration of line joining  P,Qis   −4+4t−2s,13−3t−3s,−27−2t+10sThe line  PQ is perpendicular to both lines Hence, 2−4+4t−2s+313−3t−3s−10−27−2t+10s=04−4+4t−2s−313−3t−3s−2−27−2t+10s=0It gives,s=3,t=2The point  P4,3,4is required point. Therefore,a+b+c=4+3+4=11