Because EF is a bit line in a triangular PBC,

So EF‖PC,

And PC belongs to planar PAC,

So flat packaging.

2) If the stem is PA=AB= 1, then

Because PA⊥ plane ABCD,

PA belongs to plane PAB,

So plane PAB⊥ plane ABCD,

CB⊥ intersection line AB,

So CB⊥ aircraft PAB,

And AF belong to the plane PAB,

So CB⊥AF,

Because pa = ab and f is the midpoint of PB,

So AF⊥PB,

Because CB and PB belong to planar PCB,

So AF⊥ flat printed circuit board,

Because e is on CB,

So PE is always on a flat PCB,

So where there is E, there is AF⊥PE.