MathematicsMedium249×since 2002Q2539Let d\mathrm{d}d be the distance of the point of intersection of the lines x+63=y2=z+11\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}3x+6=2y=1z+1 and x−74=y−93=z−42\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}4x−7=3y−9=2z−4 from the point (7,8,9)(7,8,9)(7,8,9). Then d2+6\mathrm{d}^2+6d2+6 is equal to :A75B78C72D69Check answerSkip