JEE Main 2013 (Offline) | 3D Geometry Question 294 | Mathematics

If the lines $${{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}$$ and $${{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}$$ are coplanar, then $$k$$ can have :

any value

exactly one value

exactly two values

exactly three values

AIEEE 2012

MCQ (Single Correct Answer)

-1

Out of Syllabus

A equation of a plane parallel to the plane $$x-2y+2z-5=0$$ and at a unit distance from the origin is :

$$x-2y+2z-3=0$$

$$x-2y+2z+1=0$$

$$x-2y+2z-1=0$$

$$x-2y+2z+5=0$$

AIEEE 2012

MCQ (Single Correct Answer)

-1

If the line $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over 4}$$ and $${{x - 3} \over 1} = {{y - k} \over 2} = {z \over 1}$$ intersect, then $$k$$ is equal to :

$$-1$$

$${2 \over 9}$$

$${9 \over 2}$$

$$0$$

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