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Question

If the lines ax+y+1=0,x+by+1=0 and x+y+c=0 (a,b,c being distinct and different from 1) are concurrent, then (11−a)+(11−b)+(11−c)=

A
0
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B
1
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C
1(a+b+c)
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D
None of these
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Solution

The correct option is C 1
The given lines are concurrent. Therefore
∣ ∣a1a2a3b1b2b3c1c2c3∣ ∣=0

∣ ∣a111b111c∣ ∣=0
Applying C2C2C1 and C3C3C1
we get
∣ ∣a1a1a1b1010c1∣ ∣
Expanding, we get
a(b1)(c1)(c1)(1a)(b1)(1a)=0
Dividing by (1a)(1b)(1c) we get
a1a+11b+11c=0
Adding 1 on both the sides we get
11a+11b+11c=1

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