We know that, three lines

$a_{1}x+b_{1}y+c_1=0$

$a_{2}x+b_{2}y+c_2=0$

$a_{3}x+b_{3}y+c_3=0$

are concurrent if

$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0$

Therefore, given lines are concurrent if

$\left|\begin{array}{ccc}1& a& a\\ b& 1& b\\ c& c& 1\end{array}\right|=0$

Expanding along $R_1$, we have

$1\left(1-bc\right)-a\left(b-bc\right)+a\left(bc-c\right)=0$

or $ab+bc+ca=1+2abc ...\left(\mathrm{i}\right)$

Now,

$\frac{a}{a-1}+\frac{b}{b-1}+\frac{c}{c-1}$

$=\frac{a\left(b-1\right)\left(c-1\right)+b\left(a-1\right)\left(c-1\right)+c\left(a-1\right)\left(b-1\right)}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}$

$=\frac{3abc+\left(a+b+c\right)-2\left(ab+bc+ca\right)}{1-\left(a+b+c\right)+\left(ab+bc+ca\right)-abc}$

Substituting the value of $ab+bc+ca$ from $\left(\mathrm{i}\right)$, we get

$\frac{3abc+\left(a+b+c\right)-2-4abc}{1-\left(a+b+c\right)+1+2abc-abc}$

$=\frac{-abc+\left(a+b+c\right)-2}{2-\left(a+b+c\right)+abc}$

$=-\left\{\frac{abc-\left(a+b+c\right)+2}{abc-\left(a+b+c\right)+2}\right\}$

$=-1$