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:
$$ pa+qb+rc=\lambda\big(p(q-r)+q(r-p)+r(p-q)\big) = \cdots $$
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[ _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 $\,u \
e 0\,$, and $\,p,q,r\,$ were all multiplied by $\,v \
e 0\,$, 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 $\,0\,$.