Find the equation of the plane passing through the line intersection of the plane: `vec2x-y=0 and 3z-y=0` and perpendicular to the plane `4x+5y-3z=8`

Options

A. `28x-17y+9z=0`
B. `28x+17y+9z=0`
C. `28x-17-9z=0`
D. `7x-3y+z=0`

Answer

A

Explanation

Equation of the plane containing the line of intersection of the planes `(2x-y)=0 and (y-3z)=0` is
`(2x-y)+lamda(y-3z)=0`
`rArr" "2x+(-1+lamda)y-3lamdaz=0` . . . (i)
Since, it is perpendicular to the plane `4x+5y-3z-8=0`, therefore
`[4xx2+5(-1+lamda)-3(-3lamda)]=0`
`[becausea_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0]`
`rArr" "8-5+5lamda+9lamda=0rArr3+14lamda=0`
`lamda=(-3)/(14)`
On sucstituting the value of `lamda` in Eq. (i), we get
`2x+(-1-(3)/(14))y-3((-3)/(14))z=0`
`rArr" "28x-17y+9z=0`
which is the required equation of the plane.