(i) Find the equation of the plane passing through the points`(2,1,0),(5,0,1)a n d(4,11)dot` (ii) If `P` s the point `(2,1,6),` then the find the point `Q` such that `P Q` is perpendicular to the plane in (i) and the midpoint of `P Q` lies on it.

Answer

`(i) x+y-2z=3 (ii) Q(6, 5, -2)`

Explanation

We know that equation of the plane passing through three points `(x_(1), y_(1), z_(1)), (x_(2), y_(2), z_(2)), (x_(3), y_(3), z_(3))` is
`" "|{:(x-x_(1),,y-y_(1),,z-z_(1)),(x_2-x_1,,y_2-y_1,,z_2-z_1),(x_3-x_1,,y_3-y_1,,z_3-z_1):}|=0`
`" "|{:(x-2,,y-1,,z-0),(5-2,,0-1,,1-0),(4-2,,1-1,,1-0):}|=0`
`" "|{:(x-2,,y-1,,z),(3,,-1,,1),(2,,0,,1):}|=0`
or `" "x+y-2z=3`

According to the question, we have to find the image of `P(2, 1, 6)` in the plane.
Let `Q` be `(alpha, beta, gamma)`. Then
`" "(alpha-2)/(1)= (beta-1)/(1)= (gamma-6)/(-2)= (-2(2+1-12-3))/(1^(2)+1^(2)+ (-2)^(2))= 4`
`rArr" "Q(alpha, beta, gamma)-= Q(6, 5, -2)`.