Find the equation of the plane passing thruogh the line of intersection of the planes x +2y + 3z = 2 and x - y + z= 3 , and at a distance `(2)/(sqrt(3))` from point `(3,1,-1).`

Options

A. `5x-11y+z=17`
B. `sqrt(2)x+y=3sqrt(2)-1`
C. `x+y+z=sqrt(3)`
D. `x-sqrt(y)=1-sqrt(2)`

Answer

A

Explanation

Key Idea
(i) Equation on plane through intersection of two planes,
i.e. `(a_(1)x+b_(1)y+c_(1)z+d_(1))+lambda`
`(a_(2)x+b_(2)y+c _(2)z+d_(2))=0`
(ii) Distance of a point `(x_(1),y_(1),z_(1))` from
`ax+by+cz+d=0`
`=(|ax_(1)+by_(1)+cz_(1)+d|)/(sqrt(a^(2)+b^(2)+c^(2)))`
Equation of plane passing through intersection of two planes `x+2y+3z=2" and "x-y+z=3` is
`(x+2y+3z-2)+lambda(x-y+z-3)=0`
`implies(1+lambda)x+(2-lambda)y+(3+lambda)z-(2+3lambda)=0`
whose distance from (3, 1, -1) is `(2)/(sqrt(3))`
`implies(|3(1+lambda)+1.(2-lambda)-1(3+lambda)-(2+3lambda)|)/(sqrt((1+lambda)^(2)+(2-lambda)^(2)+(3+lambda)^(2)))=(2)/(sqrt(3))`
`implies(|-2lambda|)/(sqrt(3lambda^(2)+4lambda+14))=(2)/(sqrt(2))`
`implies" "3lambda^(2)=3lambda^(2)+4lambda+14`
`implies" "lambda=-(7)/(2)`
`:.(1-(7)/(2))x+(2+(7)/(2))y+(3-(7)/(2))z-(2-(21)/(2))=0`
`implies" "-(5x)/(2)+(11)/(2)y-(1)/(2)z+(17)/(2)=0`
or`" "5x-11y+z-17=0`