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. 13
B. 12
C. 11
D. 8

Answer

A

Explanation

Given line is
`(x-2)/(3)=(y+1)/(4)=(z-2)/(12)=k" "("say")`
Any point on the line is
`(3k+2,4k-1,12k+2)`
This point lies on the plane
`x-y+z=5`
`:.3k+2-(4k-1)+12k+2=5`
`implies" "11k=0impliesk=0`
`:.` Intersection point is (2, -1, 2)
`:.` Distance, between points (2, -1, 2) and (-1, -5, -10)
`=sqrt((-1-2)^(2)+(-5+1)^(2)+(-10-2)^(2))`
`=sqrt(9+16+144)=13`