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,
[tex] \frac{CP'}{CD}= \frac{P'Q'}{RD} \\ \frac{2}{5} =\frac{1}{RD} \\ RD=2.5[/tex],
so R is a midlepoint of the side HD (for details see picture).