We'll say a rectangle with coordinates $(x_1,y_1,z_1,x_2,y_2,z_2)$ is of type X if it is "thin" in the $x$ dimension, i.e. if $x_1=x_2$. Rectangles of types Y and Z are similarly defined.

Define a function $count(X,Y)$ that does the following:

Select all rectangles of types X and Y and count how many pairs of rectangles there are such that at least one of them is of type X and they share a point.

The overall solution is $count(X,Y)+count(Y,Z)+count(Z,X)$.

Algorithm for $count(X,Y)$:

Imagine that the $y$ coordinate represents time. Let time flow and observe the $(x,z)$ plane representing space at some moment in time. Rectangles of type X become vertical line segments while rectangles of type Y remain rectangles.

Rectangles of type X have some positive lifetime (because $y_1<y_2$), while rectangles of type Y exist at only one instant (because $y_1=y_2$).

As time flows, three types of events can occur:

1. A type X rectangle starts
2. A type Y rectangle starts and ends immediately
3. A type X rectangle ends

We sort the events by time and process them one by one.

To count all pairs of rectangles sharing a point, such that one is of type X and another is of type Y, we need to, whenever a rectangle $(x_1,y,z_1,x_2,y,z_2)$ of type Y comes to life, answer the question "how many vertical line segments in a set intersect the rectangle $(x_1,z_1,x_2,z_2)$?".

To count all pairs of rectangles sharing a point, such that one is of type X and another is of type X, we need to, whenever a rectangle $(x_1,y,z_1,x_2,y,z_2)$ of type X comes to life, answer the question "how many vertical line segments in a set intersect the (degenerate) rectangle $(x,z_1,x,z_2)$?".

Both questions can be quickly answered if we store the starting and ending points of all vertical segments in a 2-dimensional Fenwick tree (a la IOI 2001 mobiles).

Time complexity: $\mathcal{O}(N\log N+N{\log }^{2}M)$ where $M$ is the largest allowed coordinate ($999$ in this problem).