Find one set of solutions for the following system: Find one set of solutions for the following system:
\begin{cases}
1+a^2+d^2=3+b^2+e^2=3+c^2+f^2 \\
1+ab+de=0 \\
ac+df=0 \\
bc+ef=0 \\
\end{cases}

Setting $a = -d$ and $c = f$ and $ b = -e $, we are left with the top 2 equations.
Subbing in some values, these equations become, 
$1 + 2d^2 = 3 + 2e^2 = 3 + 2c^2$ and $1 + 2de = 0$.
From this, $2de = -1 \iff d = \frac{-1}{2e}$.
Thus, we have $1 + \frac{1}{2e^2} = 3 + 2e^2 = 3 + 2c^2$.
Then, $1 = 4e^2 + 4e^4 \iff e^2 = \frac{1}{\sqrt{2}} - \frac{1}{2} \iff e = \pm\sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}}$.
Now we can solve for $d$ and $c$. But we already have $d$ and $c = e$ from inspection.
We have $$\begin{equation} \begin{aligned} e = \pm\sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}} \\ d = \frac{-1}{2e} \\ b = -e \\ a = -d \\ c = f = e\end{aligned} \end{equation} .$$
