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Question

 If ∣ ∣abcc+ba+cbcaaba+bc∣ ∣=0, then the line ax+by+c=0 passes through the fixed point which is  
  1. (1,2)
  2. (1,1)
  3. (2,1)
  4. (1,0)


Solution

The correct option is B (1,1)
Applying C1aC1 and then C1C1+bC2+cC3, and taking a2+b2+c2 common from C1, we get 
=(a2+b2+c2)a∣ ∣1bcc+b1bca1b+ac∣ ∣

R2R2R1,R3R3R1

=(a2+b2+c2)a∣ ∣1bcc+b0cab0a+bb∣ ∣

Expanding along C1
=(a2+b2+c2)a(bc+a2+ab+ac+bc)
                                                 
=(a2+b2+c2)(a+b+c)

Hence, =0a+b+c=0
Therefore, line ax+by+c=0 passes through the fixed point (1,1)

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