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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

A

13

B

12

C

11

D

8

Text Solution

Verified by Experts

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