Find the equation of the plane which is perpendicular to the plane `5x+3y+6\ z+8=0` and which contains the line of intersection of the planes `x+2y+3z-\ 4\ =\ 0\ ` and `\ 2x+y\ -\ z+5\ =\ 0`

Options

A. `51x-18y-50z+173=0`
B. `51x+18y-50z+173=0`
C. `51x+17y-50z+173=0`
D. `51x+18y-50z-173=0`

Answer

B

Explanation

The equation of a plane through the line of intersection of the plane `x+2y+3z-4=0`
and `2x+y-z+5=0` is
`(x+2y+3z-4)+lamda(2x+y-z+5)=0`
`impliesx(1+2lamda)+y(2+lamda)+z(-lamda+3)-4+5lamda=0`…………..i
Also this is perpendicular to the plane `5x+3y+6z+8=0`
`:. 5(1+2lamda)+3(2+lamda)+6(3-lamda)=0 [ :' a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0]`
`:. lamda=-29//7`
From Eq. (i)
`x[1+2((-29)/7)]+y(2-29/7)+z(29/7+3)-4+5((-29)/6)=0`
`implies x(7-58)+y(14-29)+z(29+21)-28-145=0` ltbgt `implies -51x-15y+50z-173=0`
So the required equation of plane is `51x+15y-50z+173=0`