Determine a so that the intersection line between the planes
$P_1: 2x+ay-z=3$
$P_2: x-2y+az=5$
are parallel to the plane $P_3: 2x+y+z=2$.
I want to solve this using determinants in some way.
Im thinking the intersection line between the planes $P_1$, $P_2$ is the solution the equation of system with $P_1$, $P_2$ and there are infinitely many solutions
$x=T$
$y=S$
$z=3-2T+aS$
and then this would be parallel to the plane $P_3$, can I use some determinant $=0$?
Correct answer is $a=-3/2$ or $a=3$.
Best Answer
You can directly solve for $$ \begin{vmatrix} 2 & a & -1\\ 1 & -2 & a \\ 2 & 1 & 1 \end{vmatrix}=0$$
There are various justifications for this.
First, as $P_1$ and $P_2$ have linear independent normals, there is a single line of intersection. Because that line does not pass through $P_3$, there is no solution $(x, y, z)$ to the set of three equations, hence the determinant is $0$.
Secondly, the line is perpendicular to the normals of $P_1$ and $P_2$ so it's calculated via a cross product. The line is perpendicular to the third normal as well so their dot product is 0. Put it together we have the scalar triple product is $0$.
EDIT: One should verify that the line does not lie on the third plane, because both justifications also work if the planes meet at a single line (infinite solutions).