Find the distance of the point `(-1, -5, -10)` from the point of intersection of the line `(x-2)/(3)=(y+1)/(4)=(z-2)/(12)` and plane `x-y+z=5`.

Options

A. `10`
B. `8`
C. `21`
D. `13`

Answer

D

Explanation

The coordinates of any point on the line `(x-2)/3=(y+1)/4=(z-3)/12` are given by
`(x-2)/3=(y+1)/4=(z-2)/12=r`
Suppose the line cuts the plane at `P(3r+2,4r-1,12r+2)`
This point lies on`x-y+z=5`
`:.3r+2-4r+1+12r+2=5implies11r=0impliesr=0`
Thus the line cuts the plane at `P(2,-1,2)`
The distance of `P(2,-1,2)` from `Q(-1,-5,-10)` is given by
`PQ=sqrt((-1-2)^(2)+(-5+1)^(2)+(-10-2)^(2))=13`