Show that $D$ is not linearly equivalent to any other effective divisor Let $C$ be a nonsingular quartic, $P_1,P_2,P_3 \in C$. Let $D=P_1+P_2P_3.$ Let $L$ and $L'$ be lines such that $L \bullet C= P_1+P_2+P_4+P_5$ and $L'\bullet C= P_1+P_3+P_6+P_7.$ Suppose these seven points are distinct. Show that $D$ is not linearly equivalent to any other effective divisor. Here I've followed notations provided in Fulton's algebriac curve book. 
My attempt: We know $div(LL')=\sum ord_P(LL').P=\sum_{Q\in C\cap LL'}I(Q,Q\cap LL')Q=C\bullet LL'$ nad so from the given conditions we've $div(LL')=D+P_1+P_4+P_5+P_6+P_7$. Now as $C$ is nonsingular, so multiplicity of every point of $C$ on $C$ is $1$ and hence $E=0$. So, $div(LL')=D+E+P_1+P_4+P_5+P_6+P_7$. Now if $D\equiv D'$, for some effective divisor $D'$ then by RESEDUE THEOREM exist $G'$ such that $div(G')=D'+P_1+P_4+P_5+P_6+P_7$. Now I'm stuck, please help someone. All notations are from Fulton's algebraic curve book
 A: Let us follow the hint. I assume by a nonsingular quartic you mean a plane quartic. We have
$$\operatorname{div}(LL') = D + P_1 + P_4 + P_5 + P_6 + P_7$$
as you claimed. But then, if $D \sim D'$ for some $D'$, we have
$$\operatorname{div}(C') = D' + P_1 + P_4 + P_5 + P_6 + P_7$$
for some conic $C'$. But this conic passes through the points $P_1,P_4,P_5,P_6,P_7$, and so by Bézout's theorem (§5.3), $C'$ must be equal to $LL'$ (since the number of intersections of $L$ and $C'$ is greater than $2$, so $L \subseteq C'$, and similarly $L' \subset C'$). Then, $D \sim D'$.

I wrote the solution below before realizing your question was from §8.1. Maybe it'll be helpful when you get to §8.6.
We will use the Riemann–Roch theorem (§8.6), and the fact that $L(D)$ (up to scaling by $k^\times$) corresponds to effective divisors linearly equivalent to $D$ (Exercise 8.33). We then want to show that $l(D) = 1$, and so there is a unique divisor $D$ linearly equivalent to it.
By the Riemann–Roch theorem (§8.6), we have
$$l(D) = \deg(D) + 1 - g + l(W-D) = 3 + 1 - 3 + l(W-D) = 1 + l(W-D)$$
where $W$ is the canonical divisor on the curve $C$, and $g$ is computed by Proposition 5 in §8.3:
$$g = \frac{(4-1)(4-2)}{2} = 3.$$
Now we want to show $l(W-D) = 0$. By Proposition 8 in §8.5, we have $W \sim \operatorname{div}(\tilde{L})$ is the intersection of a line with $C$. Since $\deg (W-D) = 1$, if $l(W - D) > 0$, then there is a point $Q$ such that $W - D \sim Q$, and so $W \sim D + Q = P_1 + P_2 + P_3 + Q$, that is, there exists a line $\tilde{L}$ such that its intersection with $C$ consists of the four points $P_1,P_2,P_3,Q$. But your condition on $L,L'$ being distinct lines with distinct points of intersections implies $P_1,P_2,P_3$ are not collinear, so no such $Q$ could exist. Thus, $l(W-D) = 0$.
