Let P be the image of the point (3,1,7) with respect to the plane x-y+z=3. then the equation o the plane passing through P and containing the straight 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

We know that the image of the point `(x_(1),y_(1),z_(1)` in the plane `ax+by+cz+d=0` is given by
`(x-x_(1))/a=(y-y_(1))/g=(z-z_(1))/c=(-2(ax_(1)+by_(1)+cz_(1)+d))/(a^(2)+b^(2)+c^(2))`
Let `P(x,y,z)` be the image of the point `(3,1,7)` in the plane `x-y+z=3`. Then
`(x-3)/1=(y-1)/(-1)=(z-7)/1=(-2(3-1+7-3))/(1^(2)+(-1)^(2)+1^(2))`
`impliesx=-1,y=5,z=3`
So the image of the given point is `P(-1,5,3)`.
The equation of a plane pasing through `P(-1,5,3)` is
`a(x+1)+b(y-5)+c(z-3)=0`............i
This will contain the line `x/1=y/2=z/1`, if
`a=5b-3c=0`
and `a+2b+c=0`
`:.a/(-5+6)=b/(-3-1)=c/(2+5)` or `a/1=b/(-4)=c/7`
Substituting the values of `a,b,c` in (i) we obtain `x-4y+7z=0` as the required plane.