Find the rotation of a tangent plane on a 3D sphere I've got a 3D sphere and I've one specific point (for example on location 20 meters on the X axis, 0 meters on the Y axis and -20 meters on the Z axis) on that sphere. The middle of the sphere is $\begin{pmatrix}0&0&0\\ \end{pmatrix}$ (units in meters) and got an unknown radius the radius is 
$$\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}=r$$
$$\sqrt{(20-0)^2+(0-0)^2+(-20-0)^2}=r$$
$$28.28427=r$$
On that point I'll place a 3D plane. How must I rotate that plane so it's centered the middle of the sphere?
Here is a simple drawing of the situation but must in three dimensions.


*

*Given: the point of the middle of the plane

*Asked: Degrees of the corners on the X, Y and Z axis and the plane



I would like to got a formula that could compute that. Corners must be computed in degrees.
 A: Since the question has been significantly revised, I have revised this answer.
My interpretation of the question is that the plane must pass through a given point $T$ on the sphere, and that the plane must be perpendicular to the radius of the sphere at the given point $T.$
We could also say that the plane must be tangent to the sphere at the given point $T$; that is a different way of stating the requirement, but the result will always be the same, because a tangent plane is always perpendicular to the radius at the tangent point.
You want to know the angle that the plane makes with each of the three coordinate axes. In the given example, where $T=(20,0,-20),$ the plane is parallel to the $y$ axis and does not intersect it at all.
Although there is no intersection point at which to measure the angle, it is convenient to say that the angle with the axis is zero.
In order to calculate angles, it is useful to construct a vector of length $1$ perpendicular to the plane. This vector is also known as a unit normal vector. ("Unit" because it has length $1,$ "normal" because it is perpendicular to the plane.)
The point of tangency, $T,$ defines a vector $\newcommand{r}{\mathbf r}\r$ from the center of the sphere to $T$ which is perpendicular to the tangent plane. Because the coordinates of the center are defined to be $(0,0,0)$ in this problem, the vector $\r$ will have the same coordinates as $T$;
in the example, $\r = T - (0,0,0) = (20,0,-20) - (0,0,0) = (20,0,-20) = T.$
The length of $\r$ is not $1,$ however, so we divide its length by the radius of the sphere in order to get a unit vector, and call the result $\hat\r.$

The radius of the sphere is just the length of the vector $\r,$ written
$\lVert\r\rVert,$ as implicitly observed already in the text of the question.
Therefore
$$
\hat\r = \frac{\r}{\lVert\r\rVert}
 = \frac{(20,0,-20)}{\sqrt{20^2 + 0^2 + (-20)^2}}
 = \frac{(20,0,-20)}{20\sqrt2}
 = \left(\frac1{\sqrt2}, 0, -\frac1{\sqrt2}\right).
$$

We will use the following fact about the "dot product" of two vectors
$\newcommand{u}{\mathbf u}\u$ and $\newcommand{v}{\mathbf v}\v$:
$$ \u\cdot \v = \lVert \u\rVert \lVert\v\rVert \cos\theta,$$
where $\theta$ is the angle between $\u$ and $\v.$
A unit vector parallel to the $x$ axis is 
$\newcommand{x}{\hat{\mathbf x}}\x = (1,0,0),$ so to get the angle $\theta_x$ between $\hat\r$ and $\x$ we observe that
$$ \x \cdot \hat\r = \lVert\x\rVert \lVert\hat\r\rVert \cos\theta_x
   = (1)(1)\cos\theta_x = \cos\theta_x.
$$
It follows that $\theta_x = \arccos(\x \cdot \hat\r),$
where $\arccos$ is the inverse of the cosine function.
In the example given, the left side of the equation above evaluates as
$$ \x \cdot \hat\r 
  = (1,0,0) \cdot \left(\frac1{\sqrt2}, 0, -\frac1{\sqrt2}\right)
  = \left(1\left(\frac1{\sqrt2}\right), 0(0),
       0\left(-\frac1{\sqrt2}\right)\right)
  = \frac1{\sqrt2} = \hat r_x,
$$
where $\hat r_x$ is the $x$-coordinate of $\hat\r.$
In fact, no matter where we put $T$ on the sphere, in the end
$\x \cdot \hat\r = \hat r_x,$ and therefore
$$
\theta_x = \arccos(\hat r_x).
$$
Similarly, the angles $\hat\r$ makes with the $y$-axis and with the $z$-axis
are $\theta_y = \arccos(\hat r_y)$ and $\theta_z = \arccos(\hat r_z),$
respectively, where $\hat r_y$ and $\hat r_z$ are the $y$-coordinate
and $z$-coordinate of $\hat r.$
Now we have three angles, but these are not the angles you asked for.
So consider the triangle made by the three points $M=(0,0,0)$ (the center of the sphere), $T$ (the point of tangency), and $A$ (the intersection of the plane with the $x$-axis).
These three points form a right triangle perpendicular to the tangent plane.
(For the general point $T,$ if the $y$ coordinate is not zero this triangle will not be in the $x,z$ plane, but it will still be a right triangle
and it will still be perpendicular to the tangent plane.)
We have found the angle of this triangle at $M,$ namely, $\theta_x.$
The angle at $A$ is $90^\circ - \theta_x,$ because the sum of the two angles must be $90$ degrees.
So if $\phi_x$ is the angle at which the plane intersects the $x$-axis at $A,$ then
$$ \phi_x = 90^\circ - \theta_x = 90^\circ - \arccos(\hat r_x). $$
We can write this a little bit simpler by using the fact that
$\sin(\theta) = \cos(0^\circ -\theta)$ for any angle $\theta,$ so
$$ \phi_x = \arcsin(\hat r_x) $$
where $\arcsin$ is the inverse sine function.
Similarly, the angles at which the tangent plane intersects the $y$-axis
and the $z$-axis are $\phi_y = \arcsin(\hat r_y)$ and
$\phi_z = \arcsin(\hat r_z),$ respectively.
Summary
In summary, the algorithm is:
Take the coordinates of the tangent point $T$ and divide by the radius of the sphere. (If $T = (T_x,T_y,T_z),$ the radius is
$\sqrt{T_x^2 + T_y^2 + T_z^2}.$)
The vector you get from this operation, $(\hat r_x, \hat r_y, \hat r_z),$
is a unit normal vector of the tangent plane at $T.$
The angles that the tangent plane makes with the $x$-axis, $y$-axis, and
$z$-axis are then (respectively)
$$\phi_x = \arcsin(\hat r_x),$$
$$\phi_y = \arcsin(\hat r_y),$$
$$\phi_z = \arcsin(\hat r_z).$$
If you implement this algorithm in a typical programming language using a typical math library, the $\arcsin$ function may be a function named
asin, and the output of this function will be an angle in radians.
Since you want an answer in degrees, if the function gives radians you will
need to convert them to degrees by multiplying by $\frac{\pi}{180}.$

Here are some parts of the old answer. I'm leaving them up because they might  be useful to someone trying to work with a geometry like the one in the question.
The description of the "centered" plane is very unclear.
Based on your explanation of the purpose of this plane, I guess you mean
that the line of sight from $(0,0,0)$ to $(20,0,-20)$
should be perpendicular to the plane.
In that case, the plane must have a normal vector with coordinates 
$(20,0,-20)$.
If you want the unit normal vector, multiply the coordinates of 
$(20,0,-20)$ by $\frac{1}{\sqrt{20^2 + 0^2 + 20^2}},$
so the result is $\left(\frac{1}{\sqrt2},0,-\frac{1}{\sqrt2}\right).$
The equation of a plane through the point $(20,0,-20)$
with unit normal vector $\left(\frac{1}{\sqrt2},0,-\frac{1}{\sqrt2}\right)$
is $\frac{1}{\sqrt2}x - \frac{1}{\sqrt2}z = 20\sqrt2.$
In this equation, on the left side we multiply each of the variables
$x,$ $y,$ and $z$ by the corresponding coordinate of the unit vector
(the term with $y$ disappears in this example because $0\times y = 0$), and find the correct constant value to put on the right hand side by plugging in the coordinates of a point on the plane, in this case $(20,0,-20).$
