Find the intersection of
$2x_1+x_2+x_3-3=0$, $2x_1+x_2+4x_3-6=0$, $2x_1+x_2-2=0$.
So I calculated the determinant of the three planes and it's equal to zero, then I calculated the cross product between the two first planes and it gave me $3x_1-6x_2$
The thorem in the course book says that if the determinant of the three planes is equal to zero, and cross product it's not zero, then there's two posibilities if $r_4-up_4-vq_4=0$ then the three planes intersect in a line $l$ and $l=\{(x_1,x_2,x_3):SX=up×q\}$ and S is a point in the line.
What is $r_4$, $up_4$ and $vq_4$ ? Thanks for your help.
These three planes are pairwise non-identical and non-parallel so they will meet in a point, or a line or they will form an open prism.
If you write it like $Ax=B$ and if $\det|A|=0$, the planes will either meet in a line or the will form an open prism. If you find the edges of pairwise planes by finding cross products of their normals $\vec n_1, \vec n_2, \vec n_3$. If edges are identical/coincident the planes will meet in a line. If edges are non coplanar planes will form a prism.
But more simply here if you put (3) in (1) and (2), you get $x_3=1$ and $x_3=2$ contradiction (an inconsistency). So these three Eqs. are inconsistent and hence they form a prism. In a prism two plane meet leaving out one, so no solution (inconsistency).