Question
Planes are drawn through the points (5, 0, 2) and (3, -2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelopiped so formed.

Answer

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Clearly, PBEC and QDAF are the planes parallel to the yz-plane such that their distances from the yz-plane are 5 and 3, respectively.

$\therefore$ PA = Distance between planes PBEC and QDAF

= 5 - 3

= 2

PB is the distance between planes PAFC and BDQE that are parallel to the zx-plane and are at distances 0 and -2, respectively, from the zx-plane.

$\therefore$ PB = 0 - (-2)

= 2

PC is the distance between parallel planes PBDA and CEQF that are at distances 2 and 5, respectively, from the xy-plane.

$\therefore$ PC = 2 - 5

= -3

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