Answer: 768

Small Hint:

Count the coordinate-plane crossings of the space diagonal

Big Hint:

Correct for crossings of two or three grid planes at once using greatest common divisors

Solution:

For an $a\times b\times c$ array, the diagonal crosses $a-1,$ $b-1,$ and $c-1$ internal grid planes of the three orientations. Crossings of two orientations coincide $\gcd(a,b)-1,$ $\gcd(a,c)-1,$ and $\gcd(b,c)-1$ times, and triple crossings occur $\gcd(a,b,c)-1$ times. Adding one for the initial cube and applying inclusion-exclusion gives $$\begin{gathered}a+b+c-\gcd(a,b)\\-\gcd(a,c)-\gcd(b,c)\\+\gcd(a,b,c).\end{gathered}$$ Here the pairwise gcds are $6,$ $75,$ and $3,$ and the triple gcd is $3.$ Thus the number of cube interiors met is $$\begin{gathered}150+324+375\\-6-75-3+3=768.\end{gathered}$$