Trirectangular tetrahedron: uniqueness of vertex given base Given the vertices of a triangle that forms the base of a trirectangular tetrahedron, is the right angle vertex (the vertex opposite the base where all three face angles are right angles)  uniquely defined (except, perhaps for a mirror reflection)? 
I am guessing this can be expressed in vector notation as: Given a set of three vectors ($\vec{a},\vec{b},\vec{c}$), is there only one vector $\vec{x}$ that does not lie on the plane defined by the vertices of ($\vec{a},\vec{b},\vec{c}$), and satisfies the following relation?
$$ \lvert (\vec{a}-\vec{x}) \cdot (\vec{b}-\vec{x})\lvert \ + \ 
\lvert (\vec{b}-\vec{x}) \cdot (\vec{c}-\vec{x}) \lvert \ + \ 
\lvert (\vec{c}-\vec{x}) \cdot (\vec{a}-\vec{x})\lvert \ =0\ $$
(except perhaps for the vector obtained by reflecting the vertex of $\vec{x}$ about the plane defined by ($\vec{a},\vec{b},\vec{c}$))
Apologies if this is a bit elementary. Its been a while since college geometry!
 A: The vertex is unique (up to mirror reflection in the base).
You can prove this with vector methods, but here's a more synthetic approach.

Let $PABC$ be a tri-rectangular tetrahedron with "right corner" at $P$, and let $D$, $E$, $F$ be the feet of altitudes from $A$, $B$, $C$ in $\triangle ABC$.
Consider plane $PAD$ (that is, the plane of $\triangle PAD$). Since $AD \perp BC$, we must have $PAD \perp ABC$; likewise $PBE \perp ABC$ and $PCF \perp ABC$. The three altitudes of a triangle meet at a common point called the orthocenter; consequently, our three planes meet at a common line, $\ell$, through the orthocenter, $Q$, of $\triangle ABC$, and this $\ell$ is perpendicular to $ABC$. One observes that $P$ (on $\ell$) must be a specific distance away from $ABC$[*] in order to create right angles $\angle BPC$, $\angle CPA$, $\angle APB$; as a result, $P$ is in one of two mirrored locations, as claimed. $\blacksquare$

[*] A "continuity" argument shows this: When $P$ is in plane $ABC$, clearly the angles are too large; as $P$ moves away from the plane, the angles get continually smaller until they're too small. Somewhere in there, and only once (on a given side of the plane), the angles are (ahem) just right.
For a more-constructive justification, let's write
$$a := |BC| \quad b := |CA| \quad c := |AB| \quad u := |PA| \quad v := |PB| \quad w := |PC| \qquad h := |PQ|$$
Since $\triangle PBC$, $\triangle PCA$, $\triangle PAB$ are all right triangles, we have
$$v^2 + w^2 = a^2 \qquad w^2 + u^2 = b^2 \qquad u^2 + v^2 = c^2$$
We can solve this system for $u$, $v$, $w$. The values happen to be
$$u^2 = \frac{1}{2}\left(-a^2+b^2+c^2\right) \qquad
v^2 = \frac{1}{2}\left(a^2-b^2+c^2\right) \qquad
w^2 = \frac{1}{2}\left(a^2+b^2-c^2\right)$$
but that's a bit beside the point. Now, simply note that we can compute the volume of the tetrahedron in two ways 
$$V = \frac{1}{3} h \; |\triangle ABC|= \frac{1}{6} u v w$$
Since $u$, $v$, $w$ are determined by $a$, $b$, $c$, as is area $|\triangle ABC|$, height $h$ must have a unique value.
A: The locus of points in ${\mathbb R}^3$ from which the segment $AB$ is seen under a right angle is the "Thales' sphere" with center the midpoint of $AB$ and diameter $AB$.
Therefore we have to consider  the three Thales' spheres over the three sides of the base triangle. The  two spheres over $AB$ and $AC$ intersect in a circle $\gamma$ passing through $A$. This circle intersects the sphere over $BC$ in at most two points $P_i$, which then have to lie symmetrically with respect to the base plane. From each of these two points the three sides of the base are seen under a right angle.
The circle $\gamma$ need not intersect the third sphere. In this case the problem has no solution. Consider as an example the case where the angle at $A$ is almost $180^\circ$ and $|AB|=|AC|$.
