Find coordinates of point $C$ lying on both the planes such that the perimeter of $\triangle ABC$ is minimum

Question

$A(3\mid −1\mid −1)\quad$ 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.

Answer 1

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 $\frac{dy}{dt}=\frac{1+2t}{\sqrt{2+2t+2t^2}}+\frac{2t}{\sqrt{3+2t^2}}=0$
$t=-\frac{2-\sqrt2}2$
$C\left(2\mid-\frac{2-\sqrt2}2\mid-\frac{\sqrt2}2\right)$

Answer 2

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 $C\equiv (2,t-1,-t)\equiv(2,\dfrac{1}{\sqrt{2}}-1,-\dfrac{1}{\sqrt{2}})$