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 A and B respectively. Then the distance of the mid-point of the line segment A B from the plane 2 x-2 y+z= 14 is [10-Apr-2023 shift 2]

JEE Mains 10-Apr-2023 Shift 2 Solved Paper

Section: Mathematics

Question : 20 of 90

Marks: +1, -0

Let the line ‌

=‌

6−y

=‌

z+8

intersect the lines ‌

x−5

=‌

y−7

=‌

z+2

and ‌

x+3

=‌

3−y

=‌

z−6

at the points A and B respectively. Then the distance of the mid-point of the line segment AB from the plane 2x−2y+z= 14 is

[10-Apr-2023 shift 2]

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