If the lines `a x+y+1=0,x+b y+1=0a n dx+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 \( dx + y + c = 0 \) are concurrent, then \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1 \), we can use the determinant method.

### Step-by-Step Solution:

1. **Set up the determinant for concurrency**:
The lines are concurrent if the following determinant is equal to zero:

\[
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