Given: $x\:intercept\:\:and\:\:z\:intercept$ are $3\:\:and\:\:4$ for the plane.
Let the $y\:intercept$ be $b$.
$\therefore$ Eqn. of the plane is $\large\frac{x}{3}+\frac{y}{b}+\frac{z}{4}$$=1$
$\Rightarrow\:$ The normal ($\overrightarrow n$) to the plane is $ (\large\frac{1}{3},\frac{1}{b},\frac{1}{4})$
Also it is given that the plane is parallel to $y\:axis$.
$\Rightarrow\:$ Its normal is $\perp$ to $y\:axis$
$\Rightarrow\:$ Its $y\:intercept$ is $0.$
$\therefore$ The eqn. of the plane is $\large\frac{x}{3}+\frac{z}{4}$$=1$
$or\:\:\:4x+3z=12$