Explanation
$$\begin{vmatrix}p & q& r\\q & r&p \\r & p&q\end{vmatrix}=0$$
$$p(rq-p^2)-q(q^2-rp)+r(qp-r^2)=0$$
$$3pqr-p^3-q^3-r^3=0$$
$$p^3+q^3+r^3=3pqr$$
$$p^3+q^3+r^3-3pqr=0$$
$$(p+q+r)(p^2+q^2+r^2-pa-qr-pr)=0$$
$$\Rightarrow p+q+r=0$$ or
$$p^2+q^2+r^2-pq-qr-pr=0$$
Given $$x+ay+a=0$$ ...(1)
$$bx+y+b=0$$ ...(2)
$$cx+cy+1=0$$ ...(3)
These lines are concurrent means they have only one intersection point
From 1 & 2,
$$(ab-1)y+ab-b=0$$
$$y=\dfrac{b-ab}{ab-1}$$ & $$x=-a\left(\dfrac{b-1}{ab-1}\right)$$
From 1 & 3,
$$(ac-c)y+ac-1=0$$
$$y=\dfrac{1-ac}{ac-c}$$ & $$x=-a\left(\dfrac{1-c}{ac-c}\right)$$
So,
$$\dfrac{b-ab}{ab-1}=\dfrac{1-ac}{ac-c}$$
$$\Rightarrow abc-cb+abc-a^2bc=ab-a^2bc-1+ac$$
$$\Rightarrow 1+2abc=ab+ac+bc$$ ...(4)
Now,
$$\dfrac{a}{a-1}+\dfrac{b}{b-1}+\dfrac{c}{c-1}=\dfrac{ab-a+ab-b}{ab-a-b+1}+\dfrac{c}{(c-1)}$$
$$=\dfrac{2abc-ac-bc-2ab+a+b+abc-ac-bc+c}{abc-ac-bc+c-ab+a+b-1}$$
$$=\dfrac{3abc-2(ac+bc+ab)+a+b+c}{abc-(ac+bc+ab)+a+b+c-1}$$
$$=\dfrac{abc-(ac+bc+ab)+(a+b+c)-1}{abc-(ac+bc+ab)+(a+b+c)-1}=1$$ [From (4)]
Hence, option C.
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