Find the distance of the point (1,0,-3) from plane x-y-z=9 measured parallel to the line `(x-2)/2=(y+2)/2=(z-6)/(-6)`

Explanation

The given plane is `x-y-z=9" "`(i)
The given line `AB` is `(x-2)/(2)=(y+2)/(3)=(z-6)/(-6)" "`(ii)
The equation of the line passing through `(1, 0, -3)` and parallel to `(x-2)/(2)=(y+2)/(2)=(z-6)/(-6)` is
`" "(x-1)/(2)=(y-0)/(3)=(z+3)/(-6)` =r `" "`(iii)
Coordinate of any point on (iii) may be given as `P(2r+1, 3r, -6r-3)`.
If P is the point of the intersection of (i) and (iii), then it must lie on (i). Therefore,
`" "(2r+1)-(3r)-(-6r-3)=9`
`" "2r+1-3r+6r+3=9or r=1`
Therefore, the coordinates of P are 3, 3, -9.
`" "` Distance between `Q(1, 0, -3) and P(3, 3, -9)` = `sqrt((3-1)^(2)+(3-0)^(2)+(-9+3)^(2))`
`" "=sqrt(4+9+36)=7`