Answer :

We have,

x – 2y – 3 = 0


3x + ky – 1 = 0


The given equation is in the form: a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0


Here, we have:


a1 = 1, b1 = – 2 and c1 = – 3


And, a2 = 3, b2 = k and c2 = – 1


, and


These graph lines will intersect at a unique point when we have:





Hence, k has all real values other than – 6


Thus, option B is correct

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