6. Given that $a, b, c, d$ are all non-zero real numbers, substituting $x=c, d$ into $y=x^{2}+a x$, $y$ equals 2; substituting $x=a, b$ into $y=x^{2}+c x$, $y$ equals 1. Then the maximum value of $6 a + 2 b + 3 c + 2 d$ is ( ).
(A) $6 \sqrt{3}$
(B) $\frac{8 \sqrt{3}}{3}$
(C) $\frac{10 \sqrt{3}}{3}$
(D) 0
6.A.

According to the problem, we have $\left\{\begin{array}{l}c^{2}+a c=2, \\ d^{2}+a d=2, \\ a^{2}+a c=1, \\ b^{2}+b c=1 .\end{array}\right.$
(1) + (3) gives $(a+c)^{2}=3$, i.e., $a+c= \pm \sqrt{3}$.

Thus, $a+\mathbf{c} \neq 0$.
(1) $\div$ (3) gives $c=2 a$.

Therefore, $a= \pm \frac{\sqrt{3}}{3}, c= \pm \frac{2 \sqrt{3}}{3}(a 、 c$ have the same sign $)$.
(1) - (2) gives $(c-d)(c+d+a)=0$,
(3) - (4) gives $(a-b)(a+b+c)=0$.

We consider four cases:
(1) When $a=b, c=d$,
$$
6 a+2 b+3 c+2 d=8 a+5 c= \pm 6 \sqrt{3} ;
$$
(2) When $a=b, c \neq d$, from equation (5) we get $a+c+d=0$,

then
$$
\begin{array}{l}
6 a+2 b+3 c+2 d=8 a+3 c+2(-a-c) \\
=6 a+c= \pm \frac{8 \sqrt{3}}{3} ;
\end{array}
$$
(3) When $a \neq b, c=d$, from equation (6) we get $a+b+c=0$, then
$$
\begin{array}{l}
6 a+2 b+3 c+2 d=6 a-2(a+c)+5 c \\
=4 a+3 c= \pm \frac{10 \sqrt{3}}{3} ;
\end{array}
$$
(4) When $a \neq b, c \neq d$, from equations (5) and (6) we get $a+c+d=0, a+b+c=0$, then
$$
\begin{array}{l}
6 a+2 b+3 c+2 d \\
=6 a-2(a+c)+3 c-2(a+c)=2 a-c=0 .
\end{array}
$$

In summary, the maximum value of $6 a+2 b+3 c+2 d$ is $6 \sqrt{3}$.