Please help me solving this algebraic equation I was solving a Quantative Aptitude test set and came to this question. I am not getting how to take this equation to the value it is asking for.
Please see:-
if $$ \frac{a}{q-r} = \frac{b}{r-p} = \frac{c}{p-q} $$, find the value of $$ pa+qb+rc ? $$
Options we have:-
1) 0
2) 1
3) 2
4) ­-1
Please explain your answer so that I can trick in similar kind of questions.
Thanks 
 A: Hint: let the common ratio be $\lambda\,$:
$$ \lambda = \frac{a}{q-r} = \frac{b}{r-p} = \frac{c}{p-q} $$
Then $a=\lambda(q−r),b=\lambda(r−p),c=\lambda(p−q)\,$, and:
$$
pa+qb+rc=\lambda\big(p(q-r)+q(r-p)+r(p-q)\big) = \cdots
$$

[ EDIT ]  For an alternative shortcut:  given that this was presented as a multiple choice question, and assuming the rules of the game guarantee that one of the choices must be the correct answer, then that answer can only be a) $\;=0\,$. This is because of a straightforward homogeneity argument: if $\,a,b,c\,$ were all multiplied by some constant $\,u \ne 0\,$, and $\,p,q,r\,$ were all multiplied by $\,v \ne 0\,$, then the ratios $\,a/(q-r)=b/(r-p)=c/(p-q)\,$ would still be equal, but the sum $\,pa+qb+rc\,$ would get multiplied by $\,u \cdot v\,$. But the only value among $\,\{0,1, 2, -1\}\,$ invariant to scaling is $\,0\,$.
A: Note the property: $$\frac{a}{b} = \frac{c}{d} \Rightarrow \frac{a}{b} = \frac{a+c}{b+d}.$$
Multiply the numerator and denominator by $p,q,r$, respectively, and apply the rule above: $$\frac{a}{q-r} = \frac{b}{r-p} = \frac{c}{p-q} \Rightarrow \frac{ap}{pq-pr} = \frac{bq}{qr-qp} = \frac{cr}{r(p-q)} \Rightarrow $$
$$\frac{ap}{pq-pr} = \frac{ap+bq}{r(q-p)} = \frac{cr}{r(p-q)} \Rightarrow$$
$$ap+bq=-cr \Rightarrow ap+bq+cr=0.$$
