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To find the distance between two parallel planes given by the equations:

1. \( x - 2y + 3z = 1 \)
2. \( x - 2y + 3z = -10 \)

The general formula for the distance \( d \) between two parallel planes \( Ax + By + Cz + D_1 = 0 \) and \( Ax + By + Cz + D_2 = 0 \) is:

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

Here, the planes are:
- \( x - 2y + 3z = 1 \), which can be written as \( 1x - 2y + 3z - 1 = 0 \), so \( A = 1 \), \( B = -2 \), \( C = 3 \), and \( D_1 = -1 \).
- \( x - 2y + 3z = -10 \), which can be written as \( 1x - 2y + 3z + 10 = 0 \), so \( D_2 = 10 \).

Now, applying the formula:

\[
d = \frac{| -1 - 10 |}{\sqrt{1^2 + (-2)^2 + 3^2}}
= \frac{| -11 |}{\sqrt{1 + 4 + 9}}
= \frac{11}{\sqrt{14}}
\]

Thus, the distance between the planes is:

\[
d = \frac{11}{\sqrt{14}} \approx 2.94 \, \text{units}.
\]

Would you like more details on any part of the calculation?

Here are some related questions you might find helpful:
1. How do you find the distance between a point and a plane?
2. Can the distance between non-parallel planes be calculated?
3. How does changing the coefficients in plane equations affect the distance?
4. What is the geometric interpretation of the distance between two planes?
5. How do normal vectors relate to the distance between planes?

**Tip**: Always verify that the planes are indeed parallel by checking if their normal vectors are identical.

Find the distance between the parallel planes x - 2y + 3z = 1 and x - 2y + 3z = -10.

Calculate the distance between two parallel planes x - 2y + 3z = 1 and x - 2y + 3z = -10 using the formula for the distance between parallel planes. This problem involves understanding the geometry of planes in 3D space and applying the distance formula.

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