How to solve this six variables of simultaneous equation? The problems is:
$A^2+D^2=5$
$B^2+E^2=2$
$C^2+F^2=6$
$AB+DE=3$
$BC+EF=1$
$AC+DF=1$
Please tell me what is the roots for this simultaneous equation, no process or just solve it by calculator are both fine to me.
Thank you. 
 A: There is no triple of (real) vectors in $\mathbb R^2$   with the desired Gram matrix; the angles 
There are triples of real vectors in $\mathbb R^3$ it can be done, not with integers. The quick proof is simply that the Gram matrix has determinant $5,$ not a square number. 
Finally, this arrangement exists with integer coordinates in $\mathbb R^4,$
the basis vectors are the rows of
$$
B = 
\left(
\begin{array}{cccc}
2 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 1 & 1 & 2 \\
\end{array}
\right)
$$ 
$$
G = BB^T = 
\left(
\begin{array}{ccc}
5 & 3 & 1  \\
3 & 2 & 1  \\
1 & 1 & 6  \\
\end{array}
\right)
$$
A: I obtain, for the RHS equal to $2$,
$$
A=\sqrt{2 - F^2}, \; 
B=\sqrt{2-F^2},\;
C=\sqrt{2- F^2},\;
D=E=F,
$$
or the negative square roots. So $A=B=C$ and $D=E=F$.
The edited problem has no solution (for RHS $5,2,6,3,1,1$).
A: You can reduce to three unknowns $x,y,z$ making $$A=\sqrt5\sin(x),\space D=\sqrt5\cos(x)\\B=\sqrt2\sin(y),\space E=\sqrt2\cos(y)\\C=\sqrt6\sin(z),\space F=\sqrt6\cos(z)\\AB+DE=\sqrt{10}\cos(x-y)=3\\BC+EF=\sqrt{12}\cos(y-z)=1\\AC+DF=\sqrt{30}\cos(x-z)=1$$ This gives $$\arccos\left(\dfrac{3}{\sqrt{10}}\right)=x-y\approx0.3217505\\\arccos\left(\dfrac{1}{\sqrt{12}}\right)=y-z\approx1.2779535\\\arccos\left(\dfrac{1}{\sqrt{30}}\right)=z-x\approx1.3871923$$
So you can verify that there is not solution to your system.
