= Solution
Set $g(w)=a\wp(w)+b\wp'(w)+c$. The <Weierstrass elliptic function> is doubly periodic, with a double pole at the origin modulo $\Lambda$ and no other poles. Its derivative has leading term $-2w^{-3}$. Since $b\ne0$, no lower-order term cancels this, so $g$ has a pole of order exactly three at zero and no others.
The point $\Phi(0+\Lambda)=[0,1,0]$ does not lie on the line, because its line equation evaluates to $b\ne0$. At every other point of the image, lying on the line is equivalent to $g(w)=0$. The three distinct image points therefore yield three distinct zeros modulo $\Lambda$. By the zero-pole count, these are all the zeros and each is simple. Hence its <principal divisor> on the torus is
$$
(g)=[w_1+\Lambda]+[w_2+\Lambda]+[w_3+\Lambda]-3[0+\Lambda].
$$
Applying the <elliptic divisor-sum identity> gives
$$
\boxed{w_1+w_2+w_3\in\Lambda.}
$$
This is the <period lattice> form of the <chord-and-tangent group law> on an elliptic cubic.
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