the mirror image of point `(3,1,7)` with respect to the plane `x-y+z=3` is `P`. then equation plane which is passes through the point `P` and contains the line `x/1=y/2=z/1`.

Options

A. `x+y-3z=0`
B. `3x+z=0`
C. `x-4y+7z=0`
D. `2x-y=0`

Answer

C

Explanation

Let image (x,y,z) of
`therefore (x-3)/(1)=(y-1)/(-1)=(z-7)/(1)=-2((3-1+7-3)/(1^(2)+1^(2)+1^(2)))=-4`
`P(x,y,z)=(-1,5,3)`
plane passing through p(-1,5,3)is
`a(x+1)+b(y-5)+c(z-3)=0`
also plane contains line `(x)/(1)=(y)/(2)=(z)/(1)`
`therefore (0,0,0)`satisfy
`implies a-5b-3c=0`
`and a+2b+c=0`
from (2) and (3)
pin in `(1)(X+1)-4(y-5)+7(z-3)=0`
`or x-4y+7z=0`