Given that, plane makes an intercept of 3 units on y - axis.
So, it means, plane passes through the point $\left(0,3,0\right)$ .
Thus,
$⟹\overline{\mathit{a}}=3\hat{\mathit{j}}$
Now, Further given that plane is parallel to $\mathit{x}\mathit{z}$ - plane.
So, it means y axis is normal to the surface of plane.
We know, direction ratios of $\mathit{y}$ - axis is $⟨0,1,0⟩$ .
So, normal vector to the surface of plane is given by
$⟹\overline{\mathbf{n}}=\hat{\mathrm{j}}$
Now, Required equation of plane is given by
$\overline{\mathbf{r}}\cdot \overline{\mathrm{n}}=\overline{\mathbf{a}}\cdot \overline{\mathrm{n}}$
On substituting the values, we get
$\overline{\mathit{r}}\cdot \hat{\mathit{j}}=3\hat{\mathit{j}}\cdot \hat{\mathit{j}}$
$⟹\overline{\mathit{r}}\cdot \hat{\mathit{j}}=3$
In cartesian form,
$⟹\left(\mathit{x}\hat{\mathit{i}}+\mathit{y}\hat{\mathit{j}}+\mathit{z}\hat{\mathit{k}}\right)\cdot \hat{\mathit{j}}=3$
$⟹\mathit{y}=3$
Hence, Required equation of plane is

Alternative Method:
As it is given that plane is to parallel to $\mathit{x}\mathit{z}$ plane.
So, Required equation of plane is $\mathbf{y}=\mathbf{k}$
Further given that, plane makes an intercept of 3 units on $\mathit{y}$ - axis. So, it means plane passes through $\left(0,3,0\right)$.
So,
$⟹\mathrm{k}=3$
Hence, Required equation of plane is
$⟹\mathrm{y}=3$