Find coordinates of point $C$ lying on both the planes such that the perimeter of $\triangle ABC$ is minimum $A(3\mid −1\mid −1)\quad$ is a point in the plane $x+y+z=1.$
$B(3\mid 1\mid 0)\quad$ is a point in the plane $2x−y−z=5.$
Find coördinates of point C lying on both the planes such that the perimeter of △ABC is minimum.
My Attempt: Had the question been of $2D$-geometry. I would have easily calculated using the concept of reflection.
Here we have to rotate one of the planes say $x=y+z=1$ so that it lies on the other plane. Then we need to find cooresponding co-ordinates of the given point $A(3,-1,-1)$ say $A'$on the rotated plane(which is the plane $2x-y-z=5)$.
Now ,we need to find equation of line joining $A'$ and $B$ and solve it with equation of line of intersection of the two given planes(which happens to be $\frac{x-2}{0}=\frac{y+1}{1}=\frac{z}{-1}$).
So please tell me how to find $A'$.
Minkowski In equality does give the answer but I am more interested in the Reflection principle.
How is Minkowski inequality derived.

Let $C\equiv (2,t-1,-t),A\equiv (3,-1,-1),B\equiv(3,1,0)$. Here $C$ lies on line of intersection of the two given planes.
$CA+CB=\sqrt{2}\left[\sqrt{t^2-t+1}+\sqrt{t^2-2t+\dfrac{5}{2}}\right]=\sqrt{2}\left[\sqrt{(t-\dfrac{1}{2})^2+\dfrac{3}{4}}+\sqrt{(t-1)^2+\dfrac{3}{2}}\right]$
So,now the problem gets reduced to a $2D$ problem.
Let $P\equiv (t,0),Q\equiv \left(\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right),R\equiv \left(1,\sqrt{\dfrac{3}{2}}\right)$
Now, $P$ lies on $x-$ axis and the points $Q$ and $R$ are on same side of the $x-$ axis.
$CA+CB=\sqrt{2}[PQ+PR]$.
So the problem reduces to finding $P$ such that $PQ+PR$ is minimum.
Equation of $Q'R$:
$Y-\sqrt{\dfrac{3}{2}}=\sqrt{3}(\sqrt{2}+1)(X-1)$
Solving with $Y=0$ we get $t=\dfrac{1}{\sqrt{2}}$.
Hence $C\equiv (2,t-1,-t)\equiv(2,\dfrac{1}{\sqrt{2}}-1,-\dfrac{1}{\sqrt{2}})$
