Solid ABCD in the figure below is a regular tetrahedron.
P is a point on the side AD, Q is a point on the side AB and R is a point on the side CD.
AP = 2 cm, BQ = 1 cm and CR = 3 cm.
This solid is cut by a plane which passes 3 points P, Q and R and divided into two solids.
The point that this plane cuts the side BC is set to S.
TA: UD = 2: 4 = 1: 2.
UB: BD = ①: (⑤ × 2-①) = 1: 9.
Furthermore, in Fig.2, UD : VC = 3 : 3 = 1 : 1.
Furthermore, in Fig.2, UD : VC = 3 : 3 = 1 : 1.
VC = 6 + 2/3 = 20 / 3.
BS : SC = 2/3 : 20/3 = 1 :10.
Therefore BS = 6 × 1 / (1 + 10) = 6/11 cm.
