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. `5sqrt(3)`
B. `3sqrt(10)`
C. `3sqrt(5)`
D. `10sqrt(3)`

Answer

D

Explanation

The equation of the line pasing through the point `A(1,-5,9)` and parallel to the line `x=y=z` is
`(x-1)/1=(y+5)/1=(z-9)/1`
Suppose it cuts the plane `x-y+z=5` at `P(lamda+1,lamda-5,lamda+9)`
As `P` lies on `x+y+z=5`
`:. lamda+1-lamda+5+lamda+9=5implieslamda=-10.`
So, the coordinates of `P` are `(-9,-15,-1)`.
Hence, required distance `-AP-sqrt(100+100+100)=10sqrt(3)`