To solve this problem, we need to find the volume of the given right tetrahedron. A tetrahedron is a three-dimensional figure with four triangular faces. The formula for the volume $V$ of a tetrahedron with a right-angle vertex can be expressed as:

$V = \frac{1}{3} \cdot \text{Base Area} \cdot \text{Height}$

In the provided diagram, the tetrahedron has the right-angle vertex at point $z$, with three labeled edges forming a right triangle. These labeled edges provide the base and height for our calculations.

The base triangle lies in the $xy$-plane, where:

• The edge along the $x$-axis is 100 units.
• The edge along the $y$-axis is 96 units.

These two edges form a right triangle, so the area of this triangular base is:

$\text{Base Area} = \frac{1}{2} \cdot 100 \cdot 96$

The height of the tetrahedron is along the $z$-axis, which is 128 units.

Now, we can calculate the volume: