Given that, plane makes an intercept of 3 units on y - axis.
So, it means, plane passes through the point (0,3,0) .
Thus,
⟹a=3

^

j

Now, Further given that plane is parallel to xz - plane.
So, it means y axis is normal to the surface of plane.
We know, direction ratios of y - axis is ⟨0,1,0⟩ .
So, normal vector to the surface of plane is given by
⟹n=

^

j

Now, Required equation of plane is given by
r⋅n=a⋅n
On substituting the values, we get
r⋅

^

j

=3

^

j

⋅

^

j

⟹r⋅

^

j

=3
In cartesian form,
⟹(x

^

i

+y

^

j

+z

^

k

)⋅

^

j

=3
⟹y=3
Hence, Required equation of plane is
⟹‌‌r⋅

^

j

=3‌ or ‌y=3
Alternative Method:
As it is given that plane is to parallel to xz plane.
So, Required equation of plane is y=k
Further given that, plane makes an intercept of 3 units on y - axis. So, it means plane passes through (0,3,0).
So,
⟹k=3
Hence, Required equation of plane is
⟹‌‌y=3