If planes are  drawn parallel to the coordinate planes through the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2), $ then the lengths of the edges of the parallelopiped formed are

$x_2 - x_1, y_2 -y_1, z_2 -z_1 $

Clearly, PA, PB and PC are the lengths of the edges of the parallelopiped shown in Figure.

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 = Distance between the planes PBEC and QDAF = $x_2-x_1$

PB is the distance the planes PAFC and BDQE which are parallel to zx-plane and are at distances $y_1$ and $y_2$, respectively, from zx-plane.

$∴ PB = y_2 - y_1$

Similarly, PC is the distance between the parallel planes PBDA and CEQF which are distances $z_1$ and $z_2$, respectively, from xy-plane.

$∴ PC = z_2 - z_1$