Let image of point $A$ across $x+y+z=12$ be



Then, equation of line will be
$$
\begin{aligned}
\frac{p-1}{1} & =\frac{q-2}{1}=\frac{r-3}{1}=\frac{-2(1+2+3-12)}{1^2+1^2+1^2} \\
& =\frac{12}{3}=4 \\
\frac{p-1}{1} & =4 \Rightarrow p=5 \\
\frac{q-2}{1} & =4 \Rightarrow q=6 \\
\frac{r-3}{1} & =4 \Rightarrow r=7 \\
\therefore \quad S(p, q, r) & =S(5,6,7)
\end{aligned}
$$

Now, equation of line $C S$ will be
$$
\begin{gathered}
\frac{x-5}{5-3}=\frac{y-6}{6-5}=\frac{z-7}{7-9} \\
\frac{x-5}{-2}=\frac{y-6}{-1}=\frac{z-7}{2}=\lambda \\
\therefore x=-2 \lambda+5, y=6-\lambda, z=2 \lambda+7
\end{gathered}
$$

Thus, coordinate of $B$ will be
$$
((-2 \lambda+5),(6-\lambda),(2 \lambda+7))
$$

Since, $B$ lies on given plane.
$$
\begin{gathered}
\therefore(-2 \lambda+5)+(6-\lambda)+(2 \lambda+7)=12 \\
\lambda=6
\end{gathered}
$$

$$
\begin{array}{rlrl}
& \therefore & B & =(-7,0,19) \\
& \therefore & O B & =\sqrt{(-7-0)^2+(0-0)^2+(19-0)^2} \\
& & =\sqrt{49+0+361}=\sqrt{410}
\end{array}
$$