Suppose that the Cartesian equations of your two parallel planes are $$ P_1 : a(x-x_1) + b(y-y_1) + c(z-z_1) = 0, \quad P_2: a(x-x_2)+b(y-y_2) +c(z-z_2) = 0.$$ By using the well-known formula for the distance from a point to a plane , we see that a point (x,y,z) lies between the two planes if and only if $$ |a(x-x_1) + b(y-y_1) + c(z-z_1)| + |a(x-x_2)+b(y-y_2) +c(z-z_2)| \\= |a(x_1-x_2)+b(y_1-y_2) +c(z_1-z_2)| .$$

I am glad to know that my suggestion solved your problem. It includes points on the planes. However, you can exclude these points easily: A point is between the two planes if it belongs to neither of the planes and satisfies the the provided equation: $$ a(x-x_1)+ b(y-y_1)+c(z-z_1) \ne 0 \mbox{ and } a(x-x_2)+ b(y-y_2)+c(z-z_2) \ne 0 \mbox{ and } |a(x-x_1)+ b(y-y_1)+c(z-z_1)| + |a(x-x_2)+ b(y-y_2)+c(z-z_2)| =|a(x_1-x_2)+ b(y_1-y_2)+c(z_1-z_2)|. $$ Translating the above expression into Mathematica code is straightforward.

If you represent your parallel planes with equations of the form f[x]=c1 and f[x]=c2 (with the same linear f), a point x lies between the planes exactly when f[x] lies in between the two constants c1,c2.