equation system with 6 third order equations, which is hard to solve.* We can not use the intermediate unknowns defined in (\[equ:16\]) and (\[equ:17\]) to the minimal problem, since this will result in 7 unknowns, but there are only 6 constraints. Here we introduce 3 new intermediate unknowns as $$\label{equ:24}
y_i = \left(1+{\bf{s}}^T{\bf{s}}\right)t_i, i=1,2,3.$$ Define ${{\bf{y}}=\left[y_1,y_2,y_3\right]^T}$. Then (\[equ:9\]) can be rewritten as $$\label{equ:25}
%\begin{split}
{r_g} =\frac{{{{\bf{c}}^T}{\bf{\bf{x}}} + {{\bf{a}}^T}{\bf{y}}}}{{1 + {{\bf{s}}^T}{\bf{s}}}} = 0,
%\end{split}$$ where ${\bf{x}}=\left[s_1^2,s_2^2,s_2^2,s_1s_2,s_1s_3,s_2s_3,s_1,s_2,s_3,1\right]^T$. Define ${\bar r_g}$ the numerator of (\[equ:25\]). Eliminating the denominator of (\[equ:25\]), we have $$\label{equ:26}
{\bar r_g ={{{\bf{c}}^T}{\bf{\bf{x}}} + {{\bf{a}}^T}{\bf{y}}}= 0}.$$ This is a quadratic equation in $s_1, s_2, s_3$ and a linear equation in the new unknowns $y_1, y_2, y_3$. Compared to the original equation from the numerator of (\[equ:9\]), the new one is much simpler. For the minimal problem, we have 6 effective equations as (\[equ:26\]). Let us divide the 6 equations into two groups. Each group has 3 equations as $$\label{equ:27}
{\bf{C}}_1{\bf{x}}+ {\bf{A}}_1{\bf{y}} = {\bf{0}_{3 \times 1}}, \quad
{\bf{C}}_2{\bf{x}}+ {\bf{A}}_2{\bf{y}} = {\bf{0}_{3 \times 1}},$$ We can easily calculate $\bf{y}$ using the first three equations as $$\label{equ:28}
{\bf{y}} = -{\bf{A}}_1^{-1}{\bf{C}}_1{\bf{x}}.$$ Substituting (\[equ:28\]) into the second part of (\[equ:27\]), we have $$\label{equ:29}
\begin{split}
f_i 