A plane with Miller indices ($h$ $k$ $l$) is the nearest lattice plane from the origin. To prove this, I have to show there is no lattice plane before the plane with Miller indices ($h$ $k$ $l$)
Let the primitive axes be $a$, $b$, $c$. So the plane with Miller indices ($h$ $k$ $l$) cuts the axes at $1/h$, $1/k$, $1/l$. We denote another set of axes $a^{\prime} = a/h$, $b^{\prime} = b/k$ and $c^{\prime} = c/l$.
Suppose there exists a parallel lattice plane $x$ which locates before ($h$ $k$ $l$) and intersects the axes at $1/h^{\prime}$, $1/k^{\prime}$ ,$1/l^{\prime}$. As it locates before, it must intersect $a^{\prime}$, $b^{\prime}$ and $c^{\prime}$ in same ratio. Let it cut the axes at $1/n$, $1/n$ and $1/n$ where $n$ is an integer.
As the plane $x$ is a lattice plane, it must touch a lattice point at $T = u a^{\prime} + v b^{\prime} + w c^{\prime}$, where $u$, $v$, $w$ are integers.
Now $\left( a^{\prime} - b^{\prime} \right)$ and $\left( a^{\prime} - c^{\prime} \right)$ are two vectors which are parallel to the plane $x$. For suitable $u$, $v$, and $w$, we can get a vector on axis $a^{\prime}$. So, $T + v \left( a^{\prime} - b^{\prime} \right) + w \left( a^{\prime} - c^{\prime} \right) = \left( u + v + w \right) a^{\prime}$.
This vector touches the plane $x$ but we see the plane intersects $a^{\prime}$ at $1/n$. So, $\left( u + v + w \right) a^{\prime} = a^{\prime}/n$, here $u$, $v$, $w$ and $n$ are integers... This is not possible. So the pane $x$ can not exist and ($h$ $k$ $l$) is the nearest plane.