To find the distance between two parallel planes, we can use the following formula:

$d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$

Here, the equation of a plane is generally written as $Ax + By + Cz + D = 0$, and $D_1$ and $D_2$ are the constants from the two plane equations. In our case, both planes have the same normal vector, given by $x + 6y + z$, so $A = 1$, $B = 6$, and $C = 1$.

The given planes are:

1. $x + 6y + z = 73$ (so $D_1 = -73$)
2. $x + 6y + z = 44$ (so $D_2 = -44$)

Now we apply the formula:

$d = \frac{|(-73) - (-44)|}{\sqrt{1^2 + 6^2 + 1^2}} = \frac{| -73 + 44 |}{\sqrt{1 + 36 + 1}} = \frac{| -29 |}{\sqrt{38}} = \frac{29}{\sqrt{38}}$

The distance between the two planes is:

$d = \frac{29}{\sqrt{38}} \approx 4.70$

Thus, the distance between the planes is approximately $4.70$ units.

Would you like a further breakdown of this solution or have any questions?

Here are five related questions for practice:

1. Find the distance between the planes $2x - y + 2z = 10$ and $2x - y + 2z = 5$.
2. Are the planes $x + 2y + 3z = 5$ and $2x + 4y + 6z = 10$ parallel?
3. Determine if the planes $x - 4y + z = 7$ and $2x - 8y + 2z = 10$ are parallel.
4. What is the distance between the parallel lines $3x - 2y = 5$ and $3x - 2y = 1$?
5. How do you find the angle between two planes?

Tip: When two planes are parallel, their normal vectors are proportional or identical.