# Plane parallel to intersection line using determinants

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 $$a=3$$.

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).

Indirect approach: We first find the direction numbers of the line L resulting from intersection of two planes:

$$l=\begin{vmatrix}a& -1\\-2& a\end{vmatrix}=a^2-1$$

$$m=\begin{vmatrix}-1& 2\\a& 1\end{vmatrix}=-1-2a$$

$$n=\begin{vmatrix}2& a\\1& -2\end{vmatrix}=-a-4$$

The normal of plane are $$N_p:(A=2, B=1, C=1)$$ and we must have $$L\bot N_p$$ ;the condition is that:

$$l\times A+m\times B+n\times C=0$$

which gives this equation:

$$2a^2-3a-9=0$$

which it's roots are $$a=3$$ and $$a=-\frac 32$$

Using this method you can conclude the method mentioned in other answer.