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