Given the rectangular pyramid $ABCDE$ with $A(-25, -15, 0), B(25, -15, 0), C(25, 15, 0), D(-25, 15, 0)$ and $E(0,0, 30)$, that is cut put the plane having a normal vector $(-1, -2, 3)$ and passing through $(0,0, 15)$.
The question is to find the volume of the parts of the pyramid on either side of the cutting plane.
My incomplete attempt, following the advice of Calvin Lin in his comment below, is as follows:
To calculate the volume of the upper part, first find the coordinates of points $P,Q,R,S,T$, then segment pentagon PQRST into three triangles: $\triangle SRQ , \triangle SQP, \triangle SPT $, let their areas be $A_1, A_2, A_3$ and the normal vector to the plane of $PQRST$ be $N = (-1, -2, 3)$, and let $h$ be the height of pyramid $PQRSTE$. Further let $A_4 = [ATP]$, the area of $\triangle ATP$, then the volume of the upper part is given by
$ V = \dfrac{1}{3} ( h (A_1 + A_2 + A_3 ) + 30 A_4 ) $
Note that
$ h = \dfrac{ ER \cdot N }{\| N \|} = \dfrac{EQ \cdot N}{ \| N \| } = \dfrac{ EP \cdot N }{ \|N \|} = \dfrac{ ET \cdot N}{\| N \|} = \dfrac{ ES \cdot N}{\| N \|} $
I have carried out the above computations and got the volume of the upper part to be $V_1 \approx 2930.447315$. From which I calculated the volume of the lower part by subtracting $V_1$ from the volume of the whole rectangular pyramid (which is $V_0 = \dfrac{1}{3}(50)(30)(30) $ ), and obtained the volume of the lower part as $V_2 = V_0 - V_1 \approx 12069.55268$
I look forward to your solutions to verify my answer.