Find the distance of the point `(1,-2,3)` from the plane `x-y+z=5` measured parallel to the line `x/2=y/3=z/-6`.

Options

A. 1
B. 2
C. 4
D. none of these

Answer

A

Explanation

The equation of the line pasing through `P(1,-2,3)` and parallel to the given line is
`(x-1)/2=(y+2)/3=(z-3)/(-6)`
Suppose it meets the plane `x-y+z=5` at the point `Q` given by
`(x-1)/2=(y+2)/3=(z-3)/(-3)=lamda` i.e. `(2lamda+1,3lamda-2,-6lamda+3)`
This lies on `x=y+z=5`. Therefore,
`2lamda+1-3lamda+2-6lamda+3=5implies-7lamda=-1implieslamda=1/7`
So, the coordinates of `Q` are `(9//7,=11//7,15//7)`
Hence required distance `=PQsqrt(4/49+9/49+36/49)=1`.