Find the image of the point `(1,"\ "2,"\ "3)` in the plane `"x"+2"y"+4"z"=38`

Explanation

The equation of the given plane is
`x+2y+4z=38`………….(i)
The d.r.'s of the normal to this plane are 1,2,4.
The equation of the line passing through the point P(1,2,3) and perpendicular to the given plane (i) is given by
`(x-1)/1=(y-2)/2=(z-3)/4=lambda` (say)

A general point on this line is
`(lambda+1,2lambda+2,4lambda+3)`.
For some value of `lambda`, let the point `Q(lambda+1, 2lambda+2,4lambda+3)` lie on the plane (i). Then,
`(lambda+1)+2(2lambda+2)+4(4lambda+3)=38 rArr 21 lambda=21 rArr lambda=1`.
`therefore` the coordinates of the foot of the perpendicular PQ are
`Q(1+1,2+2,4+3)`, i.e., `Q(2,4,7)`.
Let `R(x,y,z)` be the image of `P(1,2,3)` in the given plane.
Then, Q is the midpoint of PR.
`therefore (1+x)/2=2, (2+y)/2=4` and `(3+z)/2=7 rArr x=3,y=6,z=11`.
Hence, the image of `P(1,2,3)` in the given plane is `R(3,6,11)`.