Answer:
Since the prism has three rectangular faces, let's call them faces ABCD, ABEF, and CDFH. Without loss of generality, let's assume that the plane goes through points A, B, and C to slice through all three rectangular faces.
First, let's consider the intersection of the plane with face ABCD. Since the plane goes through points A, B, and C, it must contain the line segments AB and AC. Thus, the intersection of the plane with face ABCD is a triangle with vertices A, B, and some point D' on the edge CD.
Next, let's consider the intersection of the plane with face ABEF. Since the line segments AB and AC both lie entirely within face ABEF, their intersection with the plane gives line segments AB' and AC', respectively. Thus, the intersection of the plane with face ABEF is a quadrilateral with vertices A, B', E, and F.
Similarly, the intersection of the plane with face CDFH is a quadrilateral with vertices C, D', H, and F.
To visualize the resulting plane section, imagine unfolding the right triangular prism so that all its faces lie flat. Then the plane section would look something like this:
B'________E
/| / |
A /_|___/' |
| | | |
| D'- -|-----C
| / |____|/
F|/ H
The intersection of the plane with face ABCD is the triangle ABD', and the intersections with faces ABEF and CDFH are the quadrilaterals AB'EF and CD'HF, respectively.
Overall, the resulting plane section has four sides: AB' and B'E from face ABEF, EF and FD' from face CDFH, and D'A from face ABCD. It has one vertex at A and two vertices, B' and D', that lie on the edge CD. The fourth vertex is at some point E on the edge BE. The shape of the resulting plane section depends on the relative positions of points B and D on the edge CD, and point E on the edge BE.