Let for any three distinct consecutive terms a , b , c of an A.P, the lines a x + b y + c = 0 be concurrent at the point P and Q ( α , β ) be a point such that the system of equations x + y + z = 6 , 2 x + 5 y + α z = β and x + 2 y + 3 z = 4 , has infinitely many solutions. Then ( P Q ) 2 is equal to _______.

Step-by-step solution

Given, a , b , c are in A . P . ⇒ 2 b = a + c ⇒ a - 2 b + c = 0 Now, on comparing above equation with a x + b y + c = 0 we get, The line a x + b y + c = 0 passes through fixed point 1 , - 2 ∴ P = ( 1 , - 2 ) Now, we know that for infinite solution, D = D 1 = D 2 = D 3 = 0 D = 1 1 1 2 5 α 1 2 3 = 0 ⇒ α = 8 And D 1 = 6 1 1 β 5 α 4 2 3 = 0 ⇒ β = 6 So, the point Q = α , β = ( 8 , 6 ) Hence, P Q 2 = 7 2 + 8 2 = 49 + 64 = 113