Linear algebra - find all possible positions of the third corner?

Question

An equilateral triangle lies in the plane $x + y - z = 1$ and corners in points $(1, 1, 1)$ and $(2, 1, 2)$. Determine all possible positions of the third corner?

Answer 1

the length of the sides is $\sqrt{(1-2)^2 + (1-1)^2 + (1-2)^2} = \sqrt 2.$ let the third vertex is $(a, b, a+b-1.)$ then $(a-1,b-1, a+b-2)^T, (2-a, 1-b, 3-a-b)^T, (-1, 0, -1)$ they are of length $\sqrt 2$ and at an angle $120^\circ$ to each other. equating the inner product $$-1 = -(a-1)-(a+b-2)$$

the first equation gives you $$ 2a + b = 4$$ and equating the length $$(a-1)^2+(b-1)^2 + (a+b-2)^2 = 2 $$

solve the above equations gives you $a = 1, b = 2$ and $a = 2, b = 0.$ there are two third vertices $$(1, 2, 2), (2, 0, 1). $$

$\bf edit:$ here is another, may be more geometrical way, to do this. let the given vertices be $a = (1,1,1), b = (2,1,2).$ the mid pint $m$ of $ab$ is $m = (3/2, 1, 3/2)$ let the third vertex be $c.$ then $mc$ is orthogonal to $ba = (1,0,1)^T$ and $(1,1, -1),$ the normal of the plane. you can find that $(-1,2,1)^T$ satisfies both requirements. therefore $c$ can be written as $c = (3/2-t, 1+2t, 3/2+t).$ now making the length of $ac = \sqrt 2$ gives $$(1/2-t)^2 + 4t^2 + (1/2+t)^2 = 2 $$ and $t = \pm \dfrac{1}{2}$

therefore the two vertices are $$(1,2,2), (2,0,1)$$ as we found before.

Answer 2

Let $A = (1,1,1)$, $B = (2,1,2)$, so we're looking for possible coordinates of $C = (x_c, y_c, z_c)$.
$AB = \sqrt{1 + 0 + 1} = \sqrt{2}$, so we have:
$AC = AB$
$BC = AB$
$C \in (x+y-z = 1)$

and this gives us three equations:
$\sqrt{2} = \sqrt{(x_c - 1)^2 + (y_c - 1)^2 + (z_c-1)^2}$
$\sqrt{2} = \sqrt{(x_c - 2)^2 + (y_c - 1)^2 + (z_c - 2)^2)}$
$x_c + y_c - z_c = 1$

Hope you can continue by yourself from here.