What is the distance (in units) between the two planes
`3x+5y+7z=3 " and `9x+15y+21z=9` ?

Options

A. 0
B. 3
C. `(6)/(sqrt(83))`
D. 6

Explanation

To find the distance between the two planes given by the equations \(3x + 5y + 7z = 3\) and \(9x + 15y + 21z = 9\), we can follow these steps:

### Step 1: Identify the equations of the planes
The equations of the planes are:
- Plane 1: \(P_1: 3x + 5y + 7z - 3 = 0\)
- Plane 2: \(P_2: 9x + 15y + 21z - 9 = 0\)

### Step 2: Check if the planes are parallel or identical
To check if the planes are parallel or identical, we can rewrite the second plane's equation in terms of the first plane. Notice that:
\[ P_2 = 9x + 15y + 21z - 9 = 3(3x + 5y + 7z - 3) = 3P_1 \]
This shows that Plane 2 is just a scaled version of Plane 1, indicating that both planes are identical.

### Step 3: Conclusion about the distance
Since the two planes are identical, the distance between them is \(0\).

### Final Answer
The distance between the two planes is \(0\) units.

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