Find the image of plane $2x – y + z = 2$ in the plane mirror $x + 2y – z = 3$.
Could someone give the concept to find the image plane in shortest way possible.
One way which I think is that if we take $P_1:2x – y + z = 2$ and $P_2: x + 2y – z = 3$, then required plane is $P_1+ k P_2=0$, $k$ is constant Then I can find image of any point on $P_1$ with $P_2$ as mirror and then put that point in $P_1+ k P_2=0$ and get $k$ and hence required equation.
Let a plane $P'_1:ax+by+cz+d=0$ be a plane such that $P_2: x + 2y – z = 3$ bisects that plane and $P_1:2x – y + z = 2$, then the distance between every points on $P_2$ from $P_1$ and $P'_1$ is equal, that is $$\frac{ax+by+cz+d}{\sqrt{a^2+b^2+c^2}}=\frac{2x-y+z-2}{\sqrt{6}}$$ for simpler calculations, let $\dfrac{\sqrt{a^2+b^2+c^2}}{\sqrt{6}}=k$ thus $$ax+by+cz+d=k(2x-y+z-2)$$ we have to find $a$, $b$, $c$ and $d$ so select three points on $P_2$ like $$(3,0,0)~~~;~~~(0,\frac32,0)~~~;~~~(0,0,-3)$$ with substituation we get $$a-2k=1~~~;~~~b+k=2~~~;~~~c-k=-1~~~;~~~d+2k=-3$$ which are $P_2$ plane, from $$k=\dfrac{\sqrt{a^2+b^2+c^2}}{\sqrt{6}}=\dfrac{\sqrt{(2k+1)^2+(2-k)^2+(k-1)^2}}{\sqrt{6}}$$ concludes $k=3$ so $a=7$, $b=-1$, $c=2$ and $d=-9$. Finally the plane equation is $$\color{blue}{7x-y+2z-9=0}$$
I retain your notations: $P_1$ for initial plane and $P_2$ for mirror plane.
In the same vein as you, let us consider the pencil of planes defined by the two planes (the set of planes sharing the same intersection line as the initial plane and the mirror plane), i.e., the set of planes $(\Pi_k)$ with equations:
$$\tag{1}(\Pi_k): \ \ k(2 x \ – \ y + z - 2)+(x + 2y \ – z - 3)=0 \ \ \iff$$
$$(2k+1)x + (-k+2)y + (k-1) z +(-2k-3)=0$$
for any value of $k$.
Remark: we have inverted the place of $k$: in this way, the only plane not represented by equation (1) is initial plane ($P_1$).
By definition of a mirror (i.e., bissecting) plane, $k$ must be such that, for any point $M_0(x_0,y_0,z_0)$ in the mirror plane $(P_2)$,
$$\tag{2}dist(M_0,P_1) = dist(M_0,\Pi_k).$$
This condition can be written, using the point plane distance formula (http://mathworld.wolfram.com/Point-PlaneDistance.html) by squaring relationship (2):
$$\tag{3}\dfrac{(2x_0 – y_0 + z_0 - 2)^2}{2^2+1^2+1^2}=\dfrac{((2k+1)x_0 + (-k+2)y_0 + (k-1) z_0 +(-2k-3))^2}{(2k+1)^2 +(-k+2)^2 + (k-1)^2}$$
Let us take particular point $M_0(x_0,y_0,z_0)=(0,0,-3)$.
Relationship (3) becomes an equation in the single variable $k$ with unique solution $k=3$.
Plugging this value of $k$ in (1) gives plane equation:
$$7x-y+2z=9.$$