9. Given real numbers $a, b, c, d$, and $a \neq b, c \neq d$. If the equations: $a^{2}+a c=2, b^{2}+b c=2, c^{2}+a c=$ $4, d^{2}+a d=4$ all hold, then the value of $6 a+2 b+3 c+2 d$ is $\qquad$.
9.0 .
$$
\begin{array}{l}
\text { Given }\left(a^{2}+a c\right)-\left(b^{2}+b c\right)=2-2=0, \\
\left(c^{2}+a c\right)-\left(d^{2}+a d\right)=4-4=0,
\end{array}
$$

we get
$$
\begin{array}{l}
(a-b)(a+b+c)=0, \\
(c-d)(a+c+d)=0 .
\end{array}
$$

Since $a \neq b, c \neq d$, we have
$$
a+b+c=0, a+c+d=0 .
$$

Thus, $b=d=-(a+c)$.
$$
\begin{array}{l}
\text { Also, }\left(a^{2}+a c\right)+\left(c^{2}+a c\right)=2+4=6, \\
\left(a^{2}+a c\right)-\left(c^{2}+a c\right)=2-4=-2,
\end{array}
$$

we get $a+c= \pm \sqrt{6},(a-c)(a+c)=-2$.
When $a+c=\sqrt{6}$, $a-c=-\frac{\sqrt{6}}{3}$.
Solving, we get $a=\frac{\sqrt{6}}{3}, c=\frac{2 \sqrt{6}}{3}$.
When $a+c=-\sqrt{6}$, $a-c=\frac{\sqrt{6}}{3}$.
Solving, we get $a=-\frac{\sqrt{6}}{3}, c=-\frac{2 \sqrt{6}}{3}$.
Therefore, $6 a+2 b+3 c+2 d=6 a+3 c+4 b$
$$
=6 a+3 c-4(a+c)=2 a-c=0 .
$$