Three Dimensional Geometry Question 298
Question: If the points $ (1,,1,,k) $ and $ (-3,,0,,1) $ be equidistant from the plane $ 3x+4y-12z+13=0 $ ,then k =
Options:
A) 0
B) 1
C) 2
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
According to question, $ |3+4-12k+13|,=,|,-9,-12+13| $
$ \therefore ,3+4-12k+13=8\Rightarrow k=1 $ .
BETA
