For the line  $\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-4}{4}$,  which one of the following  is INCORRECT ?

 $\overrightarrow{r}=(1,1,1)+\lambda (4,6,8)$  as  $4i+6j=8k=2(2i+3j+4k)$
as  $2(2i+3j+4k).(i-2j+k)=2-6+4=0$
as point (2,3,4) satisfies the plane  $x-2y+z=0$
It is of course parallel to the plane  $x-2y+z=10$
Line  $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$  satisfied by both points (0,0,0) and (2,3,4).