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(B) Three lines $a_1x+b_1y+c_1=0$,$a_2x+b_2y+c_2=0$,and $a_3x+b_3y+c_3=0$ are concurrent if the determinant of their coefficients is zero: $\begin{vma

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(B) Three lines $a_1x+b_1y+c_1=0$,$a_2x+b_2y+c_2=0$,and $a_3x+b_3y+c_3=0$ are concurrent if the determinant of their coefficients is zero: $\begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} = 0$.Given lines are $x+ay+a=0$,$bx+y+b=0$,and $cx+cy+1=0$.The condition for concurrency is $\begin{vmatrix} 1 & a & a \ b & 1 & b \ c & c & 1 \end{vmatrix} = 0$.Expanding the determinant: $1(1-bc) - a(b-bc) + a(bc-c) = 0$.$1 - bc - ab + abc + abc - ac = 0$,which simplifies to $ab+bc+ca - 2abc = 1$.Let $S = \frac{a}{a-1} + \frac{b}{b-1} + \frac{c}{c-1}$.$S = \frac{a(b-1)(c-1) + b(a-1)(c-1) + c(a-1)(b-1)}{(a-1)(b-1)(c-1)}$.Expanding the numerator: $a(bc-b-c+1) + b(ac-a-c+1) + c(ab-a-b+1) = 3abc - 2(ab+bc+ca) + (a+b+c)$.Expanding the denominator: $(ab-a-b+1)(c-1) = abc - ab - ac + a - bc + b + c - 1 = abc - (ab+bc+ca) + (a+b+c) - 1$.Using $ab+bc+ca = 1+2abc$,the numerator becomes $3abc - 2(1+2abc) + (a+b+c) = -abc + (a+b+c) - 2$.The denominator becomes $abc - (1+2abc) + (a+b+c) - 1 = -abc + (a+b+c) - 2$.Thus,$S = \frac{-abc + (a+b+c) - 2}{-abc + (a+b+c) - 2} = 1$.

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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