The distance of the point `(1, -5, 9)` from the plane `x-y+z=5` measured along the line `x=y=z` is

Options

A. `3sqrt(10)`
B. `10sqrt(3)`
C. `10/(sqrt(3))`
D. `20/3`

Answer

B

Explanation

The equation of the line passing through `P(1,-5,9)` and parallel to the line `x/1=y/1=z/1` is
`(x-1)/1=(y+5)/1=(z-9)/1`
The coordinates of any point on this line are `Q(lamda+1,lamda-5,lamda+9)`.
If it lies on the plane `x-y+z=5`, then
`lamda+1-lamda+5+9=5implieslamda=-10`
So the coordinates of Q are `(-9,-15,-1)`
Hence required distance
`=PQ=sqrt((1+9)^(2)+(-5+15)^(2)+(9+1)^(2))=10sqrt(3)`.