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

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

where the planes are in the form \( Ax + By + Cz = D \).

From the given image, the two planes are:
1. \( x + 6y + z = 73 \)
2. \( x + 6y + z = 44 \)

In this case, the coefficients \( A \), \( B \), and \( C \) are the same for both planes, which are \( A = 1 \), \( B = 6 \), and \( C = 1 \).

Now, using the formula:

\[
d = \frac{|73 - 44|}{\sqrt{1^2 + 6^2 + 1^2}} = \frac{29}{\sqrt{1 + 36 + 1}} = \frac{29}{\sqrt{38}}
\]

Thus, the distance between the planes is:

\[
d = \frac{29}{\sqrt{38}} \approx 4.70 \text{ units}.
\]

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:
1. How do we derive the formula for the distance between two parallel planes?
2. Can we find the distance between planes that are not parallel?
3. What is the geometric interpretation of the normal vector to a plane?
4. How does this formula change if the planes are in different orientations?
5. What would happen if the two planes coincided?

**Tip:** Parallel planes always have identical normal vectors, which is why their distance calculation depends only on their constant terms.

Find the distance between the following parallel planes: x + 6y + z = 73 and x + 6y + z = 44.

Learn how to calculate the distance between two parallel planes x + 6y + z = 73 and x + 6y + z = 44 using the distance formula. The detailed solution explains key concepts such as normal vectors and parallel planes, suitable for students in Grades 10-12.

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