MathematicsMedium249×since 2002Q2444Let the foot of the perpendicular from the point (1, 2, 4) on the line x+24=y−12=z+13{{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3}4x+2=2y−1=3z+1 be P. Then the distance of P from the plane 3x+4y+12z+23=03x + 4y + 12z + 23 = 03x+4y+12z+23=0 is :A5B5013{{50} \over {13}}1350C4D6313{{63} \over {13}}1363Check answerSkip