Q: Is this value correct or should it be simplified? Given that $a\neq p$, $b\neq q$, $c\neq r$, and $\begin{vmatrix}
        p & b & c \\
        a & q & c \\
        a & b & r
        \end{vmatrix} =0$ Then find the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} $
I got the solution as $$\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} = - 2\left[  \frac{a}{p-a}+\frac{b}{q-b}+\frac{c}{r-c} \right]$$ 
Should it be simplified or is there another solution to it that is more suitable?

A: Hint: Collect terms that have a common denominator.
You have $X=2Y$.  Can you see another connection between $X$ and $Y$?

A: I got it figured out as below - 
$$\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} -  \left[  \frac{a}{p-a}+\frac{b}{q-b}+\frac{c}{r-c} \right]= -3\left[  \frac{a}{p-a}+\frac{b}{q-b}+\frac{c}{r-c} \right]$$ 
$$\frac{p-a}{p-a}+\frac{q-b}{q-b}+\frac{r-c}{r-c}= -3 \left[  \frac{a}{p-a}+\frac{b}{q-b}+\frac{c}{r-c} \right]$$ 
$$3 =   -3 \frac{a}{p-a}+\frac{b}{q-b}+\frac{c}{r-c} = $$
$$-1 =   \frac{a}{p-a}+\frac{b}{q-b}+\frac{c}{r-c} = $$
So, $$\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} = -2\left[  -1 \right] = 2$$
