A geometric question, no idea how to do it Let us have a cube with the middle point: $M = (4,3,1)$.
We know that $6(x+10) = 7(y+20) = 7z$ is one of the edge of the cube.
How to find the vertices of the cube?
I have no idea how to do this, any help? Thanks!
 A: Given:
$$\begin{gathered}
  M = (4,3,1) \hfill \\
  6\left( {x + 10} \right){\text{ }} = {\text{ }}7\left( {y + 20} \right){\text{ }} = {\text{ }}7z \hfill \\ 
\end{gathered} $$
Last equations describe planes which intersect:
$$\begin{gathered}
  6x - 7z =  - 60 \hfill \\
  y - z =  - 20 \hfill \\ 
\end{gathered} $$
This intersection is an edge. A parametrization is given by $P = (\frac{{40}}{3},0,20),w = \left( {\begin{array}{*{20}{c}}
  {\frac{7}{6}} \\ 
  1 \\ 
  1 
\end{array}} \right)$ and $$L:P + \lambda w$$
Direction vector from $L$ is normal to a plane which contains $M$.
This plane is given by:
$$7x + 6y + 6z - 52 = 0$$
Intersection with the given edge gives a point $Q$. This point is given
by 
$$Q = (4, - 8,12)$$
Now we have a lenght for
$$\overrightarrow {MQ}  = Q - M = 11\left( {\begin{array}{*{20}{c}}
  0 \\ 
  { - 1} \\ 
  1 
\end{array}} \right)$$
which is
$$\left\| {\overrightarrow {MQ} } \right\| = 11\sqrt 2 $$
And we have a diameter:
$$d = 2\left\| {\overrightarrow {MQ} } \right\| = 22\sqrt 2 $$
From this, we see length of one edge must be $22$.
Rest of cube can now be calculated straight forward.

A: Hint:
One edge of the cube is on the stright line: 
$
(x,y,z)=(-10,-20,0)+t(\dfrac{1}{6},\dfrac{1}{7},\dfrac{1}{7})
$
The distance of a point $X=(x,y,z)$ of this line from the center $M=(4,3,1)$ is:
$
d(XM)^2= (x-4)ì2+(y-3)^2+(z-1)^2=d(t)
$
Minimizing this function you find the middle point of an edge of the cube, and, if this minimum distance is $d_m$ then $2d_m$ is the diagonal of a face of the cube and $\dfrac{2d_m}{\sqrt{2}}$ is the leght of its edge.  
