Two lines x−12=y+13=z−14 and x−31=y−k2=z intersect at a point, if k is equal to
A
29
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B
12
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C
92
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D
16
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Solution
The correct option is D92 x−12=y+13=z−14=r (say) ⇒x=2r+1,y=3r−1,z=4r+1 Since, the two lines intersect. So, putting above values in second line, we get 2r+1−31=3r−1−k2=4r+11 2r−2=4r+1 ⇒r=−3/2 Also 3r−1−k=8r+2 ⇒k=−5r−3=152−3=92