Let $ax+by+cz=0$ be a line in projective space. Let the line be satisfied by two points $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$. Then we have $$a_1x+a_2y+a_3z=0$$ $$b_1x+b_2y+b_3z=0$$ This implies that $\begin{vmatrix} a_1&a_2\\b_1&b_2\end{vmatrix}=\begin{vmatrix} a_2&a_3\\b_2&b_3\end{vmatrix}=\begin{vmatrix} a_1&a_3\\b_1&b_3\end{vmatrix}=0$.
Now my textbook says that the equation of this line is $\begin{vmatrix} a_1&a_2\\b_1&b_2\end{vmatrix}x_1+\begin{vmatrix} a_2&a_3\\b_2&b_3\end{vmatrix}x_2+\begin{vmatrix} a_1&a_3\\b_1&b_3\end{vmatrix}x_3=0$.
I don't understand how this is. This relation should be satisfied by ANY $(x_1,x_2,x_3)$, and not just the point on the line, as the coefficients (the determinants) are all 0!! So you get $0.x_1+0.x_2+0.x_3=0$ for every $(x_1,x_2,x_3)$.
Am I correct?
Your claim that the three $2\times 2$ minors being all equal zero are simply not true. Since you claim $ax+by+cz=0$ to be a line, I assume you are talking about the projective plane with homogeneous coordinate $(x,y,z)$.
The two equations you gave is supposed to solve the line coordinates $(x,y,z)$, or rather the $(a,b,c)$ in your line equation $ax+by+cz=0$. The minors are not necessarily all $0$. Basically, you are looking for a vector that is perpendicular to two other vectors $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$, and therefore the results is given by cross product. The three minors pop up in the computation of the cross product.