Show that the disease of the point of intersection of the line `(x-2)/3=(y+1)/4=(z-2)/12` and the plane `(x-y+z=5)` from the point `(-1,-5,-10)` is 13 units.

Options

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

Answer

D

Explanation

Any point on the line `(x-2)/3 = (y+1)/4 = (z-2)/12=lambda` is
`(3lambda+2,4lambda-1,12lambda+2)`
This lies on x-y+z=5
`rArr 3lambda+2-4lambda+1+12lambda+2=5rArr lambda=0`
So, point is `(2,-1,2)`.
Its distance from `(-1,-5,-10)`
`=sqrt((2+1)^(2)+(-1+5)^(2)+(2+10)^(2))`
`=sqrt(9+16+144)=13`