Good question, this 3D geometry problem is definitely tricky!

We're starting with a triangular prism $ABC-DEF$. Since we're just doing ratios, let's assume that the volume of this prism is $1$. We do know that this volume is equal to (base) times (height), where base is, for example, the area of $ABC$. Now what about triangular pyramid $ABC-F$. Remember the volume of a pyramid is (1/3) times (base) times (height), where here the base and height are the same as prism $ABC-DEF$. Hence, pyramid $ABC-F$ has area $\dfrac{1}{3}$.

Now let's consider pyramid $F-PQC$. We can calculate it's volume as (1/3) times (base) times (height) where base now represents the area of $PQC$. Note that since $P$ and $Q$ are midpoints of $CE$ and $CD$, we know that $[PQC] = [CDE]/4$. But note that pyramid $F-CDE$ (with area 1/3) has the same height as pyramid $F-PQC$. Hence, their volumes are also in ratio $1:4$ so pyramid $F-PQC$ has area $\dfrac{1}{3}\div 4 = \dfrac{1}{12}$.

Side note: Notice that the reasoning we're doing here is very similar to reasoning we did in 2D (a triangle inside a parallelogram has half the area of the parallelogram, or if two triangles have the same height the ratio of their areas equals the ratio of their bases).

Hope this helps!