Find all points in space whose distances from the axes Ox…

Find all points in space whose distances from the axes $Ox$, $Oy$, $Oz$ are $a$, $b$, $c$ respectively.

Explanation of Method

Points at given distances a, b, c from the coordinate axes.

Step 1: System

\[ y^2+z^2=a^2,\qquad x^2+z^2=b^2,\qquad x^2+y^2=c^2. \]

Adding: $x^2+y^2+z^2=\tfrac12(a^2+b^2+c^2)$. Subtracting each equation from this gives

\[ x^2=\frac{-a^2+b^2+c^2}{2},\qquad y^2=\frac{a^2-b^2+c^2}{2},\qquad z^2=\frac{a^2+b^2-c^2}{2}. \]

Step 2: Solutions

\[M\left(\pm\sqrt{\tfrac{-a^2+b^2+c^2}{2}},\ \pm\sqrt{\tfrac{a^2-b^2+c^2}{2}},\ \pm\sqrt{\tfrac{a^2+b^2-c^2}{2}}\right)\]

Existence: all three radicands must be $\ge0$, i.e. no square exceeds the sum of the other two. So $a,b,c$ must be the sides of a triangle that is not obtuse (possibly right-angled or degenerate). In general there are $8$ points, one in each octant. A zero coordinate (for example $a^2=b^2+c^2$) reduces the number to $4$, and two zeros reduce it to $2$.

Check: $y^2+z^2=\tfrac{2a^2}{2}=a^2$, and similarly for $b$ and $c$ $\checkmark$

$a$ $b$ $c$ $M(x,y,z)$ $O$

Distances from $M$ to the axes: to $Ox$ it is $\sqrt{y^2+z^2}=a$ (blue), to $Oy$ $\sqrt{x^2+z^2}=b$ (red), to $Oz$ $\sqrt{x^2+y^2}=c$ (teal). Changing the signs of $x,y,z$ gives the other octants.

Result

\[\boxed{\,M\left(\pm\sqrt{\tfrac{-a^2+b^2+c^2}{2}},\ \pm\sqrt{\tfrac{a^2-b^2+c^2}{2}},\ \pm\sqrt{\tfrac{a^2+b^2-c^2}{2}}\right)\,}\]

Remark

Precision. The source calls these "the triangle inequalities". The conditions $a^2\le b^2+c^2$ etc. are stronger: they are the triangle inequalities for the squares, i.e. the triangle with sides $a,b,c$ must have no obtuse angle. For example $a=2$, $b=c=1.2$ satisfies the triangle inequality, but $x^2=\tfrac{-4+2.88}{2}\lt0$, so there is no solution.