How to find the circumcenter of a triangle and the length of the corresponding radius of the circle in $\mathbb{R}^3$ The problem is as follows:

Given the points A(1, -2, 0), B(4, 2, -5) and C(0, 0, 0), calculate the coordinates of the circumcenter of the triangle and the length of the radius (that is, the length between the circumcenter and any of the three of the vertices of the triangle).

I am attending grammar school and we are dealing with vectors. The curriculum includes basic operations with vectors, that is dot product, multiplying it with a number, adding two vectors together and subtracting them. However, the curricilum doesn't include cross product, vector equation of a line and vector equation of a plane.
I did some research and found the following formula to calculate the diameter (distance between the circumcenter and one of the vertices):
$$ \frac{abc}{\sqrt{(a + b + c)(-a + b + c)(a - b + c)(a + b - c)}}$$
I am now interested how can one derive such formula? Is it derived by using the vector dot product? I'd be pleased to get an example or explanation of deriving the above formula.
When it comes to locating the coordinates of the circumcenter, things get more complicated. So as to locate them, one must perform the following tasks:


*

*Locate the midpoint of each edge of the triangle

*Find the slope of the line for each edge of the triangle (which is quite difficult to do in R^3)



*Find the negative inverse of that slope, so you get the slope of a line parallel to a particular edge of a triangle

*Use the slope obtained in 3. step and the midpoint of the corresponding edge in 1. step to get the vector equation of the right bisector to that edge

*Use two right bisectors and locate their cross section. This point is the circumcenter of the triangle.



The above not only includes the vector equation of a line (which we weren't taught at school but it still somehow appeared to be on exam), but is also very time-consuming. The exam consists of approximately six tasks and lasts 60 minutes, therefore 10 minutes per each task and this problem itself takes just about 10 minutes.
Therefore I would like to know if there is any other, more efficient way of calculating the coordinates of the circumcenter. Is it possible to calculate it only by using vector dot product?
Thank you for the time taken.
 A: Let circumcentre $X$ have the position vector $\overrightarrow{OX}=x\hat{\textbf{i}}+y\hat{\textbf{j}}+z\hat{\textbf{k}}$. Denote $M_L$ as the mid-point of the line L.
$M_{AB}X$ is the perpendicular bisector to $AB$. Hence $\overrightarrow{M_{AB}X} \cdot \overrightarrow{AB}=0$ and similarly for the remaining two edges.
$\overrightarrow{AB}=3\hat{\textbf{i}}+4\hat{\textbf{j}}-5\hat{\textbf{k}}$ and
$\overrightarrow{BC}=-4\hat{\textbf{i}}-2\hat{\textbf{j}}+5\hat{\textbf{k}}$ and
$\overrightarrow{CA}=\hat{\textbf{i}}-2\hat{\textbf{j}}$
$\overrightarrow{M_{AB}X} = (x-5/2)\hat{\textbf{i}}+y\hat{\textbf{j}}+(z+5/2)\hat{\textbf{k}}$
$\overrightarrow{M_{BC}X} = (x-2)\hat{\textbf{i}}+(y-1)\hat{\textbf{j}}+(z+5/2)\hat{\textbf{k}}$
$\overrightarrow{M_{CA}X} = (x-1/2)\hat{\textbf{i}}+(y+1)\hat{\textbf{j}}+z\hat{\textbf{k}}$
Dotting the above respectively yields the following simultaneous equation
$$
\begin{align}
3x+4y-5z &= 20\\
-4x-2y+5z&=-45/2\\
x-2y+0z&=5/2
\end{align}
$$
The solution is $(x,y,z)=(2t+5/2, t, 2t-5/2), t\in\mathbb{R}$. ($\dagger$)
We need one more constraint upon the circumcentre, i.e. $X$ is coplanar with $A,B,C$. This can only be done by the scalar triple product.
$$\overrightarrow{CA} \cdot \overrightarrow{CB} \times\overrightarrow{CX}=
\begin{vmatrix}1 & 4 & x \\ -2 & 2 & y\\0 &-5&z \end{vmatrix}=
0$$
which gives $2x+y+2z=0$
Using the results of ($\dagger$), we thus have a unique solution of $X=(x,y,z)=(5/2, 0, -5/2)$ corresponding to $t=0$.
As for the formula for circumradius, refer to this link to see the derivation. Simply plug in the Heron's formula for expressing the area of triangle with three known edges.
A: Solution for the
given points 
$A(1, -2, 0)$,
$B(4, 2, -5)$ and 
$C(0, 0, 0) $
is much easier than the general case.
Note that the side lengths of $\triangle ABC$
\begin{align}
a=|BC|&=\sqrt{(4-0)^2+(2-0)^2+(5-0)^2}= 3\sqrt5
,\\
b=|AC|&=\sqrt{(1-0)^2+(-2-0)^2+(0-0)^2}=\sqrt5
,\\
c=|BA|&=\sqrt{(4-1)^2+(2-(-2))^2+(-5-0)^2}=5\sqrt2
.
\end{align}  
And since
\begin{align} 
a^2+b^2&=50=c^2
,
\end{align}  
$\triangle ABC$ is the right triangle,
for which it is well-known that
the circumcenter is located 
in the middle of the hypotenuse.
Hence the center and the radius 
of circumscribed circle are just
\begin{align} 
O&=\tfrac12(A+B)=(\tfrac52,0,-\tfrac52)
,\\
R&=\tfrac12\,|AB|=\frac{5\sqrt2}2
.
\end{align}
Edit
There is a formula for a general case
in terms of the apex coordinates $A,B,C$ 
and the angles 
$\alpha=\angle BAC$,
$\beta=\angle ABC$,
$\gamma=\angle BCA$:
\begin{align}
O&=\tfrac12\,\left(
\frac{\cos\alpha}{\sin\beta\sin\gamma}\cdot A
+\frac{\cos\beta}{\sin\gamma\sin\alpha}\cdot B
+\frac{\cos\gamma}{\sin\alpha\sin\beta}\cdot C
\right)
\tag{1}\label{1}
.
\end{align}
The same formula can be presented also 
in other equivalent forms like
\begin{align}
O&=\frac{R}{2\,S}\,(
a\,\cos\alpha\cdot A
+b\,\cos\beta\cdot B
+c\,\cos\gamma\cdot C
),
\end{align}
where $R$ is the circumradius and $S$ is the area 
of $\triangle ABC$.
It's easy to check that \eqref{1} works nicely
for the right triangle:
\begin{align}
\cos\gamma&=0,\quad \sin\gamma=1
,\\
\cos\alpha&=\sin\beta
,\\
\cos\beta&=\sin\alpha
\end{align}
and we get a known result as
\begin{align}
\require{cancel}
O&=\tfrac12\,\left(
\frac{\cancel{\cos\alpha}}{\cancel{\sin\beta}\cdot1}\cdot A
+\frac{\cancel{\cos\beta}}{1\cdot\cancel{\sin\alpha}}\cdot B
+\frac{0}{\sin\alpha\sin\beta}\cdot C
\right)
=
\tfrac12\,(A+B)
.
\end{align}
