A cube is dropped on the floor, and the triangular hole created has sides of lengths $68,$ $75,$ and $77.$ Find the depth of the hole. I dropped a heavy cube the other day, creating a dent in my floor. The triangular hole in the surface of the floor has sides of lengths $68,$ $75,$ and $77.$ Find the depth of the hole.

I drew a diagram, but I'm really not sure what to do next. I also tried taking some cross sections of the diagram, but that also doesn't do much. I'm really not sure how to continue from here. :(
Thanks in advance!!!
 A: First, you can calculate the sides of the tetrahedron whose apex is the deepest point of the dent.  We already know three of its sides which are $68, 75, 77$, and since we have perpendicular planes, then if these side lengths are $x, y, z$, then,
$x^2 + y^2 = 68^2 , x^2 + z^2 = 75^2 , y^2 + z^2 = 77^2 $ and these three equations solve to
$ x = 12 \sqrt{15} , y = 4 \sqrt{154} , z = 3 \sqrt{385} $
Now placing the origin of a coordinate frame at the intersection of the three perpendicular planes, with the axes along the edges, then the equation of the plane of the triangular hole is
$ \dfrac{x}{12 \sqrt{15} } + \dfrac{ y }{4 \sqrt{154}} + \dfrac{z}{3 \sqrt{385}} = 1 $
Hence the distance of the origin (which is the deepest point) to the plane of the hole is
$ d = \dfrac{1 }{\sqrt{ \dfrac{ 1 }{ 144(15) } + \dfrac{1}{16(154)} + \dfrac{1}{9(385) } }} $
And this comes to
$ d = 12 \sqrt{6} $
A: Take the bottom of the hole as the origin $O$, and the recessed edges from $O$ as the axes, with unit vectors $\pmb i$, $\pmb j$, and $\pmb k$ respectively along the edges of length $a$, $b$, and $c$. The vector $x\pmb i+y\pmb j+z\pmb k$ to the point $A$ vertically above $O$ in the plane of the floor is perpendicular to the vectors from $A$ to the points on the axes with respective vector positions $\pmb i$, $\pmb j$, and $\pmb j$ from $O$. This perpendicularity gives us three zero scalar products:
$$x(a-x)-y^2-z^2=0,$$
$$-x^2+y(b-y)-z(c-z)=0,$$
$$-x^2-y^2+z(c-z)=0.$$
These equations may be written $ax=by=cz=x^2+y^2+z^2.$ Substituting $y=(a/b)x$ and $z=(a/c)x$ gives
$$x=\frac{r^2}a,\qquad y=\frac{r^2}b\qquad z=\frac{r^2}c,$$
where
$\dfrac1{r^2}=\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}.$ In conclusion, we get $$|OA|=\sqrt{x^2+y^2+z^2}=r.$$
For $a=68$, $b=75$, and $c=78$, the corresponding value of $r$ is
$$\frac1{\sqrt{\dfrac1{68^2}+\dfrac1{75^2}+\dfrac1{78^2}}}.$$
A: Coordinate-free Solution
Consider a cuboid with sides $68$, $75$, and $78$.  Now the "hole" is a sixth of that cuboid.  Hence its volume equals $68\cdot75\cdot77/6=65\,450$.
(Let$ABCDEFGH$ be the cuboid.  The pyramid $ABCDE$ is third of the cuboid and the tetraeder $ABDE$ is half of that pyramid.)
Due to Heron, see https://en.wikipedia.org/wiki/Heron%27s_formula, (after some easy calculations) the square of the area of the triangle is $77^2\cdot75^2+68^2\cdot77^2+68^2\cdot75^2$.
Alternative solution:
Let $a$, $b$, and $c$ be the given lengths.  Then verify that the plane through $(a,0,0)$, $(0,b,c)$ and $(0,0,c)$ is given by the equation
$$
\left\langle\begin{pmatrix}
bc\\ ac\\ ab
\end{pmatrix},\vec x\right\rangle-abc=0,
$$
hence its distance from the origin is
$$
\frac{abc}{\sqrt{a^2b^2+a^2b^2+b^2c^2}}
$$
