Let's buils the intersection plane:
Point P is on AB and AP=2, then PB=3; point Q is on AE and AQ=1, then QE=4. Let P' be a point on CD such that CP'=2 and Q' be a point on the plane CDHG such that P'Q'=1 and P'Q' is perpendicular to CD. The line CQ' intersects HD at point R and the plane CPQR is intersection plane.
Consider triangles ΔCDR and ΔCP'Q', they are similar. So,
![(CP')/(CD)= (P'Q')/(RD) \\ (2)/(5) =(1)/(RD) \\ RD=2.5](https://img.qammunity.org/2019/formulas/mathematics/high-school/u0wsig2lfeai23k2c2nv9cbwoq1emarrsg.png)
,
so R is a midlepoint of the side HD (for details see picture).