One, (20 points) Given real numbers $a, b, c, d$ satisfy $2a^2 + 3c^2 = 2b^2 + 3d^2 = (ad - bc)^2 = 6$. Find the value of $\left(a^2 + \dot{b}^2\right)\left(c^2 + d^2\right)$.
Let $m=a^{2}+b^{2}, n=c^{2}+d^{2}$. Then
$$
\begin{array}{l}
2 m+3 n=2 a^{2}+2 b^{2}+3 c^{2}+3 d^{2}=12 . \\
\text { By }(2 m+3 n)^{2}=(2 m-3 n)^{2}+24 m n \geqslant 24 m n \\
\Rightarrow 12^{2} \geqslant 24 m n \\
\Rightarrow m n \leqslant 6 . \\
\text { Also } m n=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \\
=a^{2} c^{2}+b^{2} d^{2}+a^{2} d^{2}+b^{2} c^{2} \\
=(a c+b d)^{2}+(a d-b c)^{2} \\
\geqslant(a d-b c)^{2}=6,
\end{array}
$$

From equations (1) and (2), we get $m n=6$, that is
$$
\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=6 \text {. }
$$

[Note] Real numbers $a, b, c, d$ that satisfy the conditions exist and are not unique, for example,
$$
(a, b, c, d)=\left(\sqrt{2}, 1, \frac{\sqrt{6}}{3},-\frac{2 \sqrt{3}}{3}\right)
$$

is one of the solutions.