MathematicsMedium249×since 2002Q2400The distance of the point (1, 3, – 7) from the plane passing through the point (1, –1, – 1), having normal perpendicular to both the lines x−11=y+2−2=z−43{{x - 1} \over 1} = {{y + 2} \over { - 2}} = {{z - 4} \over 3}1x−1=−2y+2=3z−4 and x−22=y+1−1=z+7−1{{x - 2} \over 2} = {{y + 1} \over { - 1}} = {{z + 7} \over { - 1}}2x−2=−1y+1=−1z+7 is :A1083{{10} \over {\sqrt {83} }}8310B583{{5} \over {\sqrt {83} }}835C1074{{10} \over {\sqrt {74} }}7410D2074{{20} \over {\sqrt {74} }}7420Check answerSkip