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

Answer

B

Explanation

The equation of the line passing through the point (1,-5,9) and parallel to the line x=y=z is
`(x-1)/(1)=(y+5)/(1)=(z-9)/(1)=lamda" "` [say]
`:.x=lamda+a,y=lamda-5andz=lamda+9` lies on the plane `x-y+z=5`.
`:.` Distance `lamda+1-lamda+5+lamda+9=5`
`:." "lamda=-10`
So, the coordinate of Q is (-9,-15-1) and coordinate of P is (1,-5,9).
`:.` Distance `|PQ|=sqrt((10)^(2)+(10)^(2)+(10)^(2))=10sqrt(3)`