JEE Main Math Class 12 Three Dimensional Geometry Part 2 Questions

Since, the line

x+1

2

=

y−3

4

=

z+1

3

contains the point (-1,3,-1) and line

x+3

2

=

y+2

6

=

z−1

λ

contains the point(-3,-2,1)
Then, the distance between the plane23x−10y−2z+48=0 and the plane containing the lines= perpendicular distance of plane
23x−10y−2z+48=0 either from (-1,3,-1) or (-3,-2,1)
=|

23(−1)−10(3)−2(−1)

√(23)2+(10)2+(−2)2

| =

3

√633

It is given that distance between the planes
=

k

√633

⇒

k

√633

=

3

√633

⇒k=3