Equation of Plane involving Intersection of planes

Question

Find a plane through $A$ $(2, 1, -1)$ and perpendicular to the line of intersection of the planes $2x + y - z = 3$ and $x + 2y + z = 2$. Not too sure of what to do here, I know I need to find the point of intersection of the planes, and I found two points that I think are on the line of intersection, $B$ ($5/3$ ,$0$ ,$1/3$) and $C$ ($4/3$, $1/3$, $0$ ). However, I don't know where to go from there, and with the plane needing to be perpendicular. Answer given is $x$-$y$+$z$=$0$ Any help is appreciated.

Answer 1

The sum of the two planes equations gives

$$x+y=\frac 53$$

thus $$y=\frac 53-x$$

and

$$z=2-x-2y=2-\frac{10}{3}+x$$

the line intersection of the two planes has parametric equations :

$$x=0+x$$ $$y=\frac 53-x$$ $$z=\frac{-4}{3}+x$$

the vector director of this line is $$(1,-1,1)$$

the plane perpendicular to this line has the equation $$x-y+z=d$$ it contains the point $(2,1,-1)$ if $$d=0$$

the equation you are looking for is $$\boxed{x-y+z=0}$$

Answer 2

Hint:

A directing vector of the intersection line of the two planes is the cross-product $u=\vec n_1\times \vec n_2$ of the normal vectors of the planes.

Once you calculated its coordinates, the equation of the plane you're after is, in vector form $$\vec u\cdot \overrightarrow{OM}=\vec u\cdot \overrightarrow{OA}.$$