Show that P(a,b,c) , Q(b,c,a) and R(c,a,b) form an…

Show that $P(a,b,c)$, $Q(b,c,a)$ and $R(c,a,b)$ form an equilateral triangle, and find the plane containing them.

Explanation of Method

Cyclic permutations of the coordinates give an equilateral triangle in a plane perpendicular to the main diagonal.

Step 1: Equal sides

Each squared side is a sum of the same three squares, in a different order: \[|PQ|^2=|QR|^2=|RP|^2=(a-b)^2+(b-c)^2+(c-a)^2 .\]

So the triangle is equilateral, unless $a=b=c$, when the three points coincide.

Step 2: The plane

\[ \overrightarrow{PQ}\times\overrightarrow{PR}=\begin{vmatrix}\vec i&\vec j&\vec k\\ b-a&c-b&a-c\\ c-a&a-b&b-c\end{vmatrix}=-(a^2+b^2+c^2-ab-bc-ca)\{1,1,1\}. \]

The factor is nonzero unless $a=b=c$, because

\[ a^2+b^2+c^2-ab-bc-ca=\tfrac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right]. \]

So the normal is $\{1,1,1\}$.

\[x+y+z=a+b+c\]

Check: each point has coordinate sum $a+b+c$. $\checkmark$

The centroid is $G\left(\tfrac{a+b+c}3,\tfrac{a+b+c}3,\tfrac{a+b+c}3\right)$, on the diagonal $x=y=z$. The map $(x,y,z)\mapsto(y,z,x)$ is a $120^\circ$ rotation about that diagonal and sends $P\to Q\to R\to P$. This explains why the triangle is equilateral.

$P(3,1,0)$ $Q(1,0,3)$ $R(0,3,1)$ $G$ $x=y=z$

Example $a=3,\ b=1,\ c=0$: the cyclic permutations give an equilateral triangle in $x+y+z=4$. Its centroid $G$ lies on the diagonal $x=y=z$, which is the normal $\{1,1,1\}$.

Result

\[\boxed{\,x+y+z=a+b+c\,}\]