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.

The third point is
$C(2\mid t\mid -1-t).$
Using the distance formula, calculate $\overline{AC}\text{ then }\overline{BC}.$
Their sum is a function in t. $\sqrt{2+2t+2t^2}+\sqrt{3+2t^2}=y$
Calculate $\frac{dy}{dt}=\frac{1+2t}{\sqrt{2+2t+2t^2}}+\frac{2t}{\sqrt{3+2t^2}}=0$
$t=-\frac{2-\sqrt2}2$
$C\left(2\mid-\frac{2-\sqrt2}2\mid-\frac{\sqrt2}2\right)$
