一,(20 points) Given that $a, b, c, d$ are all non-zero real numbers. Substituting $x=a$ and $b$ into $y=x^{2}+c x$, the value of $y$ is 1 in both cases; substituting $x=c$ and $d$ into $y=x^{2}+a x$, the value of $y$ is 3 in both cases. Try to find the value of $6 a+2 b+3 c+2 d$.
Given the problem, we have
$$
\left\{\begin{array}{l}
a^{2}+a c=1, \\
b^{2}+b c=1, \\
c^{2}+a c=3, \\
d^{2}+a d=3 .
\end{array}\right.
$$
(1) + (3) gives $(a+c)^{2}=4 \Rightarrow a+c= \pm 2$.

Thus, we know $a+c \neq 0$.
(3) $\div$ (1) gives $c=3 a$,

so $\left\{\begin{array}{l}a=\frac{1}{2}, \\ c=\frac{3}{2}\end{array}\right.$ or $\left\{\begin{array}{l}a=-\frac{1}{2}, \\ c=-\frac{3}{2} .\end{array}\right.$
From (2) - (1) we get $(a-b)(a+b+c)=0$,
From (4) - (3) we get $(d-c)(a+d+c)=0$.
$1^{\circ}$ When $a=b, c=d$, we have
$6 a+2 b+3 c+2 d=8 a+5 c= \pm \frac{23}{2}$.
$2^{\circ}$ When $a=b, c \neq d$, from (6) we know $a+d+c=0$, so $6 a+2 b+3 c+2 d=8 a+3 c+2(-a-c)= \pm \frac{9}{2}$.
$3^{\circ}$ When $a \neq b, c=d$, from (5) we know $a+b+c=0$, so $6 a+2 b+3 c+2 d=6 a+2(-a-c)+5 c= \pm \frac{13}{2}$.
$4^{\circ}$ When $a \neq b, c \neq d$, from (5) and (6) we know $a+b+c=0$, $a+c+d=0$, so
$$
\begin{array}{l}
6 a+2 b+3 c+2 d \\
=6 a+2(-a-c)+3 c+2(-a-c)= \pm \frac{1}{2} .
\end{array}
$$