To find the equation of the plane in the image, we need to utilize the information given:

• The plane passes through the point $(3, 1, 2)$.
• It is parallel to the $xz$-plane.

Key observations:

1. A plane that is parallel to the $xz$-plane implies that the normal vector to the plane will be perpendicular to the $xz$-plane. Therefore, the normal vector will point in the direction of the $y$-axis.
2. Hence, the equation of such a plane will only involve the $y$-coordinate as a constant.

Since the plane is parallel to the $xz$-plane, the $y$-coordinate will remain fixed at $y = 1$, as the plane passes through the point $(3, 1, 2)$.

Equation of the plane:

Thus, the equation of the plane is: $y = 1$

This indicates that the plane lies at $y = 1$ for all $x$ and $z$-values.

Would you like more details on the reasoning or any further clarification? Here are some related questions you might find interesting:

1. What is the general equation of a plane given a point and a normal vector?
2. How do you determine the normal vector to a plane based on its orientation?
3. Can you explain how planes parallel to the coordinate planes are formed?
4. What changes when a plane is parallel to the $xy$- or $yz$-plane?
5. How can we find the distance between two parallel planes?

Tip: When a plane is parallel to a coordinate plane, its equation simplifies by excluding the variable in the direction perpendicular to that coordinate plane.