@CPhill

We need to write the equation of a plane, here, Julius.....

Since the prism is bisected  by the xy plane,

Let D  = (0, 0 ,-3 )

Let  A  = (0, 0, 3)

Let E  = (4, 0,3)

Let G   =(4,0,-3)

We can form the following  vectors  DG  = (4,0, 0)    and  DE  = (4, 0, 6)

[These aren't the only two possibilities......we just need any two vectors formed by the points]

Note that all the points lie in the xz  plane....so...we need a vector that is orthagonal to this plane

The cross-product of the two vectors will be orthagonal to both of these vectors...so we have

                         i          j           k       i         j

DG X DE   =    4         0          0       4        0

                        4        0           6       4        0       

 ( [ 0 * 6]i   + [ 0 * 4 ] j  + [4 * 0] k   -  (  [ 4 * 0 ] k  + [ 0 *0] i  +  [ 6 * 4]j )   =

0i   + 0 j   +  0 k   - 0 k - 0i  - 24 j

So....we have

0i  -24j - 0k       =  { 0, -24, 0 }    this is the vector  that is orthagonal to both vectors ....[and hence,  the plane containing those vectors]

And  the equation of the plane becomes

0 (x - 4) -24 (y - 0)  + 0 (z - 3)   =  0

-24y    =  0