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

Explanation

Let PN be the perpendicular from `P(1,2,3)` to the plane `x+2y+4z = 38`.
`:.` Equation of PN
`(x-1)/(1) = (y-2)/(2) = (z-3)/(4) = lambda`
(Say)
`rArr` Co-ordinates of any point on line `PN(lambda+1,2lambda+2,4lambda+3)`

Let this point be N.
This point N lies on the plane `x + 2y + 4z = 38`.
`:. (lambda + 1) + 2 (2lambda+2)+4(4lambda+3) = 38`
`rArr lambda + 1 + 4lambda+4+16lambda+12 = 38`
`rArr lambda = 1`
`:.` Co-ordinates of `N = (2,4,7)`.
Let `Q(x_(1),y_(1),z_(1))` be the image of point `P(1,2,3)`. Then N will be the mid-point of PQ.
`:. (x_(1)+1)/(2) = 2, (y_(1)+2)/(2) = 4, (z_(1)+3)/(2) = 7`
`rArr x_(1) = 3, y_(1) = 6, z_(1) = 11`
`:.` Required image `= (3,6,11)`.