Let for any three distinct consecutive terms a, b, c of an A.P, the lines ax+by+c=0 be concurrent at the point P and Q( , ) be a point such that the system of equations x+y+z=6, 2x+5y+ z= and x+2y+3z=4, has infinitely many solutions. Then (PQ)2 is equal to .
Answer
113
Solution
2b=a+c a(1)+b(-2)+c=0 every line passes through P(1, -2)
Infinitely many solutions D= vmatrix1&1&1\\2&5& \\1&2&3 vmatrix=8- =0 =8
eqn (2) =-(eqn 1) +3(eqn 3) on the left: (2, 5, 8)=-(1, 1, 1)+3(1, 2, 3)
=-6+3(4)=6 Q(8, 6)
(PQ)2=72+82=113
Answer: 113