MathematicsHard249×since 2002Q2536Let P\mathrm{P}P be the point of intersection of the lines x−21=y−45=z−21\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}1x−2=5y−4=1z−2 and x−32=y−23=z−32\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}2x−3=3y−2=2z−3. Then, the shortest distance of P\mathrm{P}P from the line 4x=2y=z4 x=2 y=z4x=2y=z isA3147\frac{3 \sqrt{14}}{7}7314B5147\frac{5 \sqrt{14}}{7}7514C147\frac{\sqrt{14}}{7}714D6147\frac{6 \sqrt{14}}{7}7614Check answerSkip