Let the line x/1=(6-y)/2=(z+8)/5 intersect the lines (x-5)/4=(y-7)/3=(z+2)/1 and (x+3)/6=(3-y)/3=(z-6)/1 at the points A and B respectively. Then the distance of the mid-point of the line segment AB from the plane 2 x-2 y+z=14 is:

Three Dimensional Geometry: JEE Main 2023 Mathematics question

JEE Main 2023 · 10 Apr, Evening shift · Q2614

JEE Main 202310 Apr, Evening shiftMathematicsMediumThree Dimensional Geometry249× since 2002

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 :