Given four points `A(2, 1, 0), B(1, 0, 1), C(3, 0, 1) and D(0, 0, 2)`. Point D lies on a line L orthogonal to the plane determined by the points A, B and C.
The equation of the plane ABC is

Options

A. `x+y+z-3=0`
B. `y+z-1=0`
C. `x+z-1=0`
D. `2y+z-1=0`

Answer

b

Explanation

`" "|{:(x-2,,y-1,,z),(1-2,,0-1,,1-0),(3-2,,0-1,,1-0):}|=0`
`" "(x-2)[(-1)-(-1)]-(y-1)[(-1)-1]+z[1+1]=0`
or `" "2(y-1)+2z=0`
or `" "y+z-1=0`
The vector normal to the plane is `vecr= 0hati+hatj+ hatk`
The equation of the line through `(0, 0, 2)` and parallel to `vecn` is `vecr = 2hatk+lamda(hatj+hatk)`
The perpendicular distance of `D(0, 0, 2)` from plane
`ABC` is `|(2-1)/(sqrt(1^(2)+1^(2))|=(1)/(sqrt2)`