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(A) The given equations of the lines are:$x+ay+a=0$$bx+y+b=0$$cx+cy+1=0$Since these lines are concurrent,the determinant of their coefficients must be

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(A) The given equations of the lines are:$x+ay+a=0$$bx+y+b=0$$cx+cy+1=0$Since these lines are concurrent,the determinant of their coefficients must be zero:$\begin{vmatrix} 1 & a & a \ b & 1 & b \ c & c & 1 \end{vmatrix} = 0$Expanding the determinant:$1(1-bc) - a(b-bc) + a(bc-c) = 0$$1 - bc - ab + abc + abc - ac = 0$$1 + 2abc = ab + bc + ac$Now,we evaluate the expression $S = \frac{a}{a-1} + \frac{b}{b-1} + \frac{c}{c-1}$.$S = \frac{a(b-1)(c-1) + b(a-1)(c-1) + c(a-1)(b-1)}{(a-1)(b-1)(c-1)}$$S = \frac{a(bc-b-c+1) + b(ac-a-c+1) + c(ab-a-b+1)}{abc - (ab+bc+ac) + (a+b+c) - 1}$$S = \frac{abc - ab - ac + a + abc - ab - bc + b + abc - ac - bc + c}{abc - (ab+bc+ac) + (a+b+c) - 1}$$S = \frac{3abc - 2(ab+bc+ac) + (a+b+c)}{abc - (ab+bc+ac) + (a+b+c) - 1}$Substitute $ab+bc+ac = 1+2abc$:$S = \frac{3abc - 2(1+2abc) + (a+b+c)}{abc - (1+2abc) + (a+b+c) - 1}$$S = \frac{3abc - 2 - 4abc + a+b+c}{abc - 1 - 2abc + a+b+c - 1}$$S = \frac{-abc + a+b+c - 2}{-abc + a+b+c - 2} = 1$

Let $a, b$ and $c$ be distinct and none of them is equal to $1$. If the lines $x+ay+a=0$,$bx+y+b=0$ and $cx+cy+1=0$ are concurrent,then the value of $\frac{a}{a-1}+\frac{b}{b-1}+\frac{c}{c-1}$ is

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