When I was solving the Senary Quadratic Diophantine Equation (1), I found the following two sets of parametric solutions consisting of quadratic forms. I have a hunch that each solution can be represented by a quadratic form in six variables, but I don't know how the second set of solutions can be represented by a quadratic form in six variables. $$\left\{ \begin{split} u&=t_1^2+t_2^2-t_3^2-t_4^2+t_5^2+t_6^2\\ v&=2 \left(t_1 t_3-t_2 t_4\right)\\ w&=2 \left(t_1 t_4+t_2 t_3\right)\\ x&=t_1^2+t_2^2+t_3^2+t_4^2-t_5^2-t_6^2\\ y&=2 \left(t_1 t_5-t_2 t_6\right)\\ z&=2 \left(t_1 t_6+t_2 t_5\right)\\ \end{split} \right.$$
The above parameter solution satisfies the equation $u^{2}+v^{2}+w^{2}=x^{2}+y^{2}+z^{2}$ (1).
$$\left\{ \begin{split} U&=T_1^2+T_2^2+T_3^2-T_4^2-T_5^2\\ V&=T_1^2+T_2^2+T_3^2+T_4^2-T_5^2-2 T_4\left(T_1+T_2+T_3-T_5\right)\\ W&=T_1^2+T_2^2+T_3^2-T_4^2+T_5^2-2 T_5\left(T_1+T_2+T_3-T_4\right)\\ X&=T_1^2-T_2^2-T_3^2+T_4^2+T_5^2+2 T_1\left(T_2+T_3-T_4-T_5\right)\\ Y&=T_1^2-T_2^2+T_3^2-T_4^2-T_5^2-2 T_2\left(T_1+T_3-T_4-T_5\right)\\ Z&=T_1^2+T_2^2-T_3^2-T_4^2-T_5^2-2 T_3\left(T_1+T_2-T_4-T_5\right)\\ \end{split} \right.$$
The above parameter solution satisfies the equation $U^{2}+V^{2}+W^{2}=X^{2}+Y^{2}+Z^{2}$.
I was wondering if there exists a set of parametric solutions consisting of quadratic forms in six variables such that the degenerate form of the parametric solution is as given by us.
In addition, is there any relevant information about the quadratic parameter solution? Thanks in advance!