Interplanar distance given Miller indice of the planes

Say we've a cubic crystal of unity edge length.


A set of planes in such a crystal is specified by their miller indice as $(3 2 0)$. One of these planes then has intercepts on the axes as $(2 ,3 ,\infty)$.
It's well known that the interplanar distance between two adjacent planes in a cubic lattice is given


$d\_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}}$,in our case this gives us


$d\_{320} = 1/\sqrt13$.


Now if I look at the planes which form the Miller indices of $ 3 2 0$ I find that apart from a plane having intercepts as $(2 ,3,0)$ an adjacent parallel plane to this plane having the same indices will pass via origin $0,0,0$ and the distance between them is
$d\_{320} = 6/\sqrt13$


Which contradicts the above result!
Can anyone please point out what's wrong in my reasoning?




Your adjacent plane isn't.
[![miller indices](https://i.stack.imgur.com/Ms8gB.png)](https://i.stack.imgur.com/Ms8gB.png)


Worth a thousand words, they say.


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