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\ a n d\ 2x+y\ -\ z+5\ =\ 0`

Options

A. `15x+15y-20z+4=0`
B. `51x+15y-50z+173=0`
C. `3x-5y+7=0`
D. `3x+5y-5z+9=0`

Answer

B

Explanation

The equation of a plane through the line of intersection of the planes.
`x+2y+3z-4=0` and `2x+y-z+5=0` is given by
`(x+2y+3z-4)+lamda(2x+y-z+5)=0`
`impliesx(1+2lamda)+y(2+lamda)+z(3-lamda)-4+5lamda=0` ………………..i
This is perpendicular to the plane `5x+3y+6z+8=0`
`:.5(1+2lamda)+3(2+lamda)+6(3-lamda)=0`
`implies7 lamda+29=0implieslamda=-29//7`
Putting `lamda=-29//7` in i we obtain the equation of the required plane as
`-51x-15y+50z-173=0implies51x+15y-50z+173=0`