Box in the Air

Let (x , y, z) be the centroid of the cuboid. Since the points (2 , 2 , 2), (7 , 5 , 2), & (7 , 2 , 4) are all centroids of the cuboid's faces, the distances from (x , y , z) to these points depict half of the cuboid's width, length, and height, as shown in the picture below.

The diagram portrays the blue centroids of the cuboid's faces with the red centroid of the cuboid.

Now considering the points (2 , 2 , 2) and (7 , 5 , 2), they have the same z-value = 2 and, therefore, are on the same x-y plane.

Then for the points (2 , 2 , 2) and (7 , 2 , 4), they have the same y-value = 2, and, therefore, are on the same x-z plane.

Finally, the points (7 , 2 , 4) and ( 7 , 5 , 2) have the same x-value = 7, so they are on the same y-z plane.

As a result, the three planes are parallel to each pair of the cuboid's faces. In other words, the centroid (x , y , z) will have the same values of x, y, & z according to the same planes of the faces's centroid: z = 2, y = 2, and x = 7, as shown above.

Now the distances from the cuboid's centroid to other centroids can be simply calculated as followed:

The distance between points (7 , 2 , 2) & (2 , 2 , 2) = $\sqrt{(7 - 2)^2 + (2 - 2)^2 + (2 - 2)^2}$ = $7 - 2 = 5$

Or simply the distance between (7 , 2 , 2) & (7 , 5 , 2) = $5 - 2 = 3$ .

The distance between (7 , 2 , 2) & (7 , 2 , 4) = $4 - 2 = 2$ .

Now remember that these distances are half of the actual width, length, and height of the cuboid, so we have to double them before applying the volume formula:

Volume of the cuboid = $(2\times 2)(2\times 3)(2\times 5) = 240$ .