Find the equation of the plane containing the lines 2x-y+z-3=0,3x+y+z=5 and a t a distance of `1/sqrt6` from the point (2,1,-1).

Answer

`62x+ 29y+ 19z-105=0`

Explanation

The given line is `2x-y+z-3=0=3x+y+z-5`, which is intersection of the following two planes :
`" "2x-y+z-3=0" "` (i)
`" " 3x+y+z-5=0" "` (ii)
Any plane containing this line will be the plane passing through the intersection of planes (i) and (ii).
Thus, the plane containing given line can be written as follows :
`" "(2x-y+z-3)+lamda(3x+y+z-5)=0`
`" "(3lamda+2)x+ (lamda-1)y+ (lamda +1)z+ (-5lamda-3)=0`
As its distance from the point `(2, 1, -1)` is `1//sqrt6`,
`" "|((3lamda+2)2+(lamda-1)1+ (lamda+1)(-1)+ (-5lamda-3))/(sqrt((3lamda+2)^(2)+ (lamda-1)^(2)+ (lamda+1)^(2)))|`
`=(1)/(sqrt6)`
`|(lamda-1)/(sqrt(11lamda^(2)+ 12lamda+6))|= (1)/(sqrt6)`
Squaring both sides, we get
`" "((lamda-1)^(2))/(11lamda^(2)+ 12lamda+6)= (1)/(6)`
or `" "5lamda^(2)+ 24lamda=0`
or `" "lamda(5lamda+24)=0`
or `" "lamda=0, -24//5`
Therefore, the required equations of planes are `2x-y+z-3=0` and
`" "[3((-24)/(5))+2]x+[ - (24)/(5) - 1] y + [-(24)/(5)+1]z- 5((-24)/(5))-3=0`
or, `" "62x+ 29y+ 19z-105=0`