If the lines `a x+y+1=0,x+b y+1=0 and x+y+c=0(a ,b ,c` being distinct and different from `1)` are concurrent, then prove that `1/(1-a)+1/(1-b)+1/(1-c)=1.`

Explanation

To prove that if the lines \( ax + y + 1 = 0 \), \( x + by + 1 = 0 \), and \( x + y + c = 0 \) (where \( a, b, c \) are distinct and different from 1) are concurrent, then \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1 \), we will follow these steps:

### Step 1: Write the equations in standard form
The equations of the lines can be rewritten as:
1. \( ax + y + 1 = 0 \) → Coefficients: \( (a, 1, 1) \)
2. \( x + by + 1 = 0 \) → Coefficients: \( (1, b, 1) \)
3. \( x + y + c = 0 \) → Coefficients: \( (1, 1, c) \)

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