Principal divisor $P+Q+R-3\infty$ on elliptic curve comes from a straight line Let $E=V(y^2-(x-a)(x-b)(x-c))$ be an elliptic curve. Let $D=P+Q+R-3\infty$ be a divisor. Then $D$ is principal if and only if $P$, $Q$ and $R$ lie on a straight line.
One direction is straightforward - if the three points $P$, $Q$, $R$ are collinear, the line on which they lie gives the divisor. But what about the other direction?
The field of definition is assumed to be algebraically closed.
Edit: None of the suggestions so far have been satisfying. The hint my professor gave us was to use Riemann-Roch.
In particular: What happens if $\infty\in\left\{P,Q,R\right\}$?
 A: Assume $D=P+Q+R-3P_\infty$ is principal. Let $R’$ be the third point of intersection of the line through $P$ and $Q$ with the curve (with usual conventions concerning multiplicities). We want to show $R=R’$. By the direction you’ve proved, $D’=P+Q+R’ - 3P_\infty$ is principal. If $R \neq R’$, then the difference $D-D’=R-R’$ is non-zero and principal, i.e. the two points $R$ and $R’$ are linearly equivalent, and this only happens when the curve is rational (so if its smooth and complete it’s a $\mathbb{P}^1$ (Hartshorne II.6.10.1 but here is a proof sketch: if $div(f)=R-R’$, consider the map $F:E \to \mathbb{P}^1$ given by $[f,1]$ and observe it has degree $1$ map of smooth curves so is an isomorphism.)
A: 
There is an argument specific to elliptic curves in Weierstrass form, it doesn't generalize well to other curves.

Let $E:y^2=x^3-ax-b, 4a^3-27b^2\ne 0$ be a smooth affine cubic curve defined over an algebraized closed field $k$. Its field of rational functions is $k(x)[y]/(y^2-x^3+ax+b)$. The projective closure is $C:ZY^2=X^3+aXZ^2+bZ^3$ which is smooth too at the only one new point $O=[0:1:0]\in C-E$ which corresponds to $(\infty,\infty)$ on $E$.

That $E$ is smooth implies that its coordinate ring $k[E]=k[x,y]/(y^2-x^3-ax-b)$ is the full ring of rational functions $C\to k$ regular on $E$.

If $D=\sum_{j=1}^n P_j- nO$ is principal $=Div(f)$ then $f$ is regular on $E$ so $f\in k[E]$ ie. $f = \sum_{j,l} c_{j,l} x^j y^l$.
Near $O$ the equation becomes $\frac{y^2}{x^3}=1+a/x^2+b/x^3$ where $1+a/x^2+b/x^3$ is regular and non-zero at $O$. Thus $x$ has a pole of order $2k$ and $y$ has a pole of order $3k$ at $O$. Since $O$ is a smooth point of $C$ then $k=1$ and $x/y$ has a simple zero.
Replacing in $f = \sum_{j,l} c_{j,l} x^j y^l$ each occurrence of $y^2$ by $x^3+ax+b$ we get $$f = g(x)+yh(x),\qquad g,h\in k[x]$$ Then $g(x)$ has a pole of order $2\deg(g)$ and $yh(x)$ has a pole of order $3+2\deg(h)$ at $O$, thus $f$ has a pole of order $n=\max(2\deg(g),3+\deg(h))$ at $O$.
And hence $n=3$ implies that $\deg(g)\le 1,\deg(h)=0$ which means that $f = c+dx+ey,e \in k^*$, so that $Div(c+dx+ey)=P_1+P_2+P_3-3O$ and $P_1,P_2,P_3$ lie on the line $c+dx+ey=0$.
Conversely if $P_1,P_2,P_3$ are 3 distinct points of $E$ lying on the line $c+dx+ey=0,e\in k^*$ then $Div(c+dx+ey)=P_1+P_2+P_3-3O$ 
