0 &  -\frac{1}{2}(w_3 + \lambda) &  w_1 &  0 &  -z_2 &  0 &  -z_1 &  \frac{1}{2}x_1 &  \frac{1}{2}y_1\\
  \frac{1}{2}(w_3 + \lambda) &  0 &  -w_2 &  z_1 &  0 &  z_2 &  0 &  -\frac{1}{2}y_2 &  -\frac{1}{2}x_2\\
  -w_1 &   w_2 &  0 &  -x_1 &  x_2 &  -y_1 &  y_2 &  0 &  0\\
  0 &  -z_1 &  x_1 &  0 &  0 &  0 &  0 &  0 &  0\\
  z_2 &  0 &  -x_2 &  0 &  0 &  0 &  0 &  0 &  0\\
  0 &  -z_2 &  y_1 &  0 &  0 &  0 &  0 &  0 &  0\\
  z_1 &  0 &  -y_2 &  0 &  0 &  0 &  0 &  0 &  0\\
  -\frac{1}{2}x_1 &  \frac{1}{2}y_2 &  0 &  0 &  0 &  0 &  0 &  0 & 0\\
  -\frac{1}{2}y_1 &  \frac{1}{2}x_2 &  0 &  0 &  0 &  0 &  0 &  0 &  0
  \end{array}\right).$$ Using variables we suppose that the intensities of the forces are made orthogonal according to the procedure described in [@KhRCD; @KhHMJ]. It is known that if two orthogonal fields ${{\boldsymbol \alpha}},{{\boldsymbol \beta}}$ (${{\boldsymbol \alpha}}{\bdot} {{\boldsymbol \beta}}= 0$) have equal intensities $a=b$, then 