= Solution
For representatives $p=(a,b,c)$ and $p'=(a',b',c')$, a point $[x:y:z]$ belongs to their <projective line> exactly when these three vectors are <linearly dependent>. The <determinant> criterion gives
$$
\begin{vmatrix}x&y&z\\a&b&c\\a'&b'&c'\end{vmatrix}=0.
$$
Expanding along the first row gives
$$
\boxed{(bc'-cb')x+(ca'-ac')y+(ab'-ba')z=0.}
$$
The coefficient vector is the <cross product> $p\times p'$. It is nonzero because the <projective points> are distinct. Rescaling either representative merely rescales all three coefficients, leaving the <projective line> unchanged. This is the join formula in <projective joins and meets by cross products>.
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