Answer :

The given lines are


ax + 12y + 1 = 0 … (1)


bx + 13y + 1 = 0 … (2)


cx + 14y + 1 = 0 … (3)


It is given that (1), (2) and (3) are concurrent.



a(13 – 14) – 12(b – c) + 14b – 13c = 0


-a – 12b + 12c + 14b – 13c = 0


-a + 2b – c = 0


2b = a + c


Hence, a, b and c are in AP.

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