The equation of the plane passes through (2, 3, 4) and parallel to the plane $x+2y+4z=5$ is :

Let a, b, c be the direction ratios of normal plane, which is passing through the point (2, 3, 4) is given as:

$a(x-2)+b(y-3)+c(z-4)=0$

And this plane parallel to the plane

$$\begin{gathered} x+2y+4z=5 \\ \begin{array}{l}\therefore \frac{\mathrm{a}}{1}=\frac{\mathrm{b}}{2}=\frac{\mathrm{c}}{4}=\mathrm{k}\\ \Rightarrow \mathrm{a}=\mathrm{k},\mathrm{b}=2\mathrm{k},\mathrm{c}=4\mathrm{k}\end{array} \end{gathered}$$

Thus the equation of the plane is

$\begin{array}{l}\mathrm{k}(\mathrm{x}-2)+2\mathrm{k}(\mathrm{y}-3)+4\mathrm{k}(\mathrm{z}-4)=0\\ \Rightarrow (\mathrm{x}-2)+2(\mathrm{y}-3)+4(\mathrm{z}-4)=0\\ \Rightarrow \mathrm{x}+2\mathrm{y}+4\mathrm{z}-24=0\end{array}$