$\def\1{x_1}\def\2{x_2}\def\3{x_3}\def\f{f(\1,\2,\3)}\def\bs{\bigskip}\def\b{\begin{pmatrix}}\def\e{ \end{pmatrix}}\def\le{\left}\def\ri{\right}\def\g{g(\1,\2,\3)}\def\l{\lambda}\def\n{\nabla}\def\x{\lambda_1}\def\y{\lambda_2}$
I am putting in the following input:
Solve[{2 + 2 a*d + 2 a*e == 0, 1 - 2 b + 2 d*b + 2 e*b == 0, 1 + 2 d*c + 2 e*c == 0, -2 + a^2 - 2 b + b^2 + c^2 == 0, -2 + a^2 + b^2 + c^2 == 0}, {a, b, c, d, e}]
and it is returning: Output[10]= {}
What am I doing wrong? Is this not computable?
$$
\begin{equation} 2+2\x\1+2\y\1 = 0%8 \end{equation}
\begin{equation} 1+2\x\2+2\y\2-2\y = 0%9 \end{equation}
\begin{equation} 1+2\x\3+2\y\3=0%10 \end{equation}
\begin{equation} \1^2 + \2^2 - 2\2+\3^2-2=0 %11 \end{equation}
\begin{equation} \1^2 + \2^2 + \3^2 - 2 = 0%12 \end{equation}$$
Using the notations of your Mathematica command, you should notice that variables $a,b,c$ can be eliminated using the first three equations. This leads to $$a=\frac{-1}{d+e}$$ $$b=-\frac{1}{2 (d+e-1)}$$ $$c=\frac{1}{2 (d+e)}$$ Now, replacing these in the next equations, the fourth becomes $$ \frac{5}{4 (d+e)^2}+\frac{1}{d+e-1}+\frac{1}{4 (d+e-1)^2}-2=0$$ and the fifth becomes $$\frac{5}{(d+e)^2}+\frac{1}{(d+e-1)^2}-8=0$$ So, only $(d+e)$ terms everywhere and then the problem.