find that the distance 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

Options

A. 13
B. 12
C. 11
D. 8

Explanation

Given line is `( x-2)/( 3) = ( y + 1)/(4) = (z -2)/( 12) k` (say ) Any point on the lines is `( 3 k + 2, 4k -1, 12k + 2 ) `
This point lies on the plane `x - y + z = 5 `
`:. 3k + 2 - (4k - 1)+ 12k+ 2 = 5 rArr 11k =0 rArr k = 0 `
`:. ` Intersection point ( 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`