Solution
= Solution
The direction <vector> is $b-a=(1,2,-2)$. The vector equation and symmetric Cartesian equation of the line are
$$
\boxed{r=(1,2,4)+t(1,2,-2),\qquad x-1=\frac{y-2}{2}=\frac{z-4}{-2}.}
$$
The $(x,y)$-plane has $z=0$, which gives $t=2$. Thus its intersection with the line has position
$$
\boxed{r=(3,6,0).}
$$