Find the distance between the parallel planes `2x-y+3z+40` and `6x-3y+9z-3=0`.

Explanation

To find the distance between the parallel planes given by the equations \(2x - y + 3z + 40 = 0\) and \(6x - 3y + 9z - 3 = 0\), we can follow these steps:

### Step 1: Verify that the planes are parallel
To check if the planes are parallel, we need to compare their normal vectors. The normal vector of the first plane \(2x - y + 3z + 40 = 0\) is \(\mathbf{n_1} = (2, -1, 3)\), and the normal vector of the second plane \(6x - 3y + 9z - 3 = 0\) is \(\mathbf{n_2} = (6, -3, 9)\).

To see if the planes are parallel, we check if \(\mathbf{n_2}\) is a scalar multiple of \(\mathbf{n_1}\):
\[
\mathbf{n_2} = 3 \cdot \mathbf{n_1}
...