Let the line … intersect the lines … and … at the points … and … respectively. Then the distance of the mid-point of the line segment … from the plane … is :

Option C. We have, … intersect the line

3d Geometry

JEE Main 2023 · 10 Apr, Evening shift · Q2614

MathematicsMedium249×since 2002Q2614

Let the line

\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}

intersect the lines

\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}

and

\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}

at the points

\mathrm{A}

and

\mathrm{B}

respectively. Then the distance of the mid-point of the line segment

\mathrm{AB}

from the plane

2 x-2 y+z=14

is :