If planes are drawn parallel to the coordinate 

planes through the points $P\left(x_1,y_1,z_1\right)\text{ and }Q\left(x_2,y_2,z_2\right),$

then the lengths of the edges of the  parallelopiped formed are 

Clearly, $PA, PB and PC$ are the lengths of the 

edges of the parallelopiped shown in Fig. 5

Clearly, $PBEC, QDAF$ are  planes parallel to $yz$-plane

 such that their distances from $yz$-plane are $X_1 and X_2$

respectively. So,

$PA= Dis\tan ce between the planes PBEC and QDAF$= $=x_2-x_1$

$PB$ is the distance between the planes $PAFC$ and $BDQE$ which 

are parallel to$zx$-plane and are at distances $y_1\text{ and }{y}_2$

respectively,  from $zx$-plane. 

$\therefore PB=y_2-y_1$

Similarly, $PC$ is the distance between the parallel planes $PBDA$

and  $CEQF$which are at distances $z_1\text{ and }{z}_2\text{, }$ respectively, from

$xy-plane.$

$\therefore PC=z_2-z_1$