about to only mathematics

Answer

`2x-y+z-3=0;62x+29y+19z-105=0`

Explanation

Equation of plane containing the lines
`2x-y+z-3=0" and "3x+y+z=5` is
`(2x-y+z-3)+lambda(3x+y+z-5)=0`
`implies(2+3lambda)x+(lambda-1)y+(lambda+1)z-3-5lambda=0`
Since, distance of plane from (2,1,-1) to above plane is `1sqrt(6).`
`:.|(6lambda+4+lambda-1-lambda-1-3-5lambda)/(sqrt((3lambda+2)^(2)+(lambda-1)^(2)+(lambda+1)^(2)))
|=(1)/(sqrt(6))`
`implies" "6(lambda-1)^(2)=11lambda^(2)+12lambda+6`
`implies" "lambda=0,-(24)/(5)`
`:.` Equations of planes are
`2x-y+z-3=0" and "62x+29y+19z-105=01`