Show that the planes $ACG$ and $BEH$ defined by the vertices of the cube shown in Figure  are parallel. What is their distance if the edge length of the cube is $1$ meter?
[img]https://cdn.artofproblemsolving.com/attachments/c/9/21585f6c462e4289161b4a29f8805c3f63ff3e.png[/img]
1. **Identify the vertices of the cube:**
   Given the vertices of the cube, we have:
   - \( A(1,0,0) \)
   - \( B(1,1,0) \)
   - \( C(0,1,0) \)
   - \( D(0,0,0) \)
   - \( E(1,1,1) \)
   - \( F(0,1,1) \)
   - \( G(0,0,1) \)
   - \( H(1,0,1) \)

2. **Find the equation of the plane \(ACG\):**
   - Points on the plane \(ACG\) are \(A(1,0,0)\), \(C(0,1,0)\), and \(G(0,0,1)\).
   - To find the equation of the plane, we use the general form \(ax + by + cz = d\).
   - Substitute the points into the plane equation:
     \[
     \begin{cases}
     a(1) + b(0) + c(0) = d \implies a = d \\
     a(0) + b(1) + c(0) = d \implies b = d \\
     a(0) + b(0) + c(1) = d \implies c = d
     \end{cases}
     \]
   - Since \(a = b = c = d\), we can choose \(d = 1\) for simplicity, giving us the plane equation:
     \[
     x + y + z = 1
     \]

3. **Find the equation of the plane \(BEH\):**
   - Points on the plane \(BEH\) are \(B(1,1,0)\), \(E(1,1,1)\), and \(H(1,0,1)\).
   - Substitute the points into the plane equation:
     \[
     \begin{cases}
     a(1) + b(1) + c(0) = d \implies a + b = d \\
     a(1) + b(1) + c(1) = d \implies a + b + c = d \\
     a(1) + b(0) + c(1) = d \implies a + c = d
     \end{cases}
     \]
   - From the first equation, \(a + b = d\).
   - From the second equation, \(a + b + c = d\).
   - From the third equation, \(a + c = d\).
   - Subtract the first equation from the second:
     \[
     (a + b + c) - (a + b) = d - d \implies c = 0
     \]
   - Subtract the third equation from the first:
     \[
     (a + b) - (a + c) = d - d \implies b = c
     \]
   - Since \(c = 0\), \(b = 0\), and \(a = d\), we can choose \(d = 2\) for simplicity, giving us the plane equation:
     \[
     x + y + z = 2
     \]

4. **Verify that the planes are parallel:**
   - The normal vector to the plane \(x + y + z = 1\) is \(\vec{n_1} = (1, 1, 1)\).
   - The normal vector to the plane \(x + y + z = 2\) is \(\vec{n_2} = (1, 1, 1)\).
   - Since \(\vec{n_1} = \vec{n_2}\), the planes are parallel.

5. **Calculate the distance between the parallel planes:**
   - The distance \(d\) between two parallel planes \(ax + by + cz = d_1\) and \(ax + by + cz = d_2\) is given by:
     \[
     d = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}
     \]
   - For the planes \(x + y + z = 1\) and \(x + y + z = 2\):
     \[
     d = \frac{|2 - 1|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{1}{\sqrt{3}}
     \]

The final answer is \(\boxed{\frac{1}{\sqrt{3}}}\).