What is the distance between the planes `x-2y+z-1=0 and -3x+6yy-3z+2=0`?

Options

A. 3 unit
B. 1 unit
C. 0
D. None of the above

Explanation

To find the distance between the two planes given by the equations \(x - 2y + z - 1 = 0\) and \(-3x + 6y - 3z + 2 = 0\), we can follow these steps:

### Step 1: Identify the coefficients of the plane equations
The first plane can be written in the form \(a_1x + b_1y + c_1z + d_1 = 0\):
- For the first plane \(x - 2y + z - 1 = 0\):
- \(a_1 = 1\)
- \(b_1 = -2\)
- \(c_1 = 1\)
- \(d_1 = -1\)

The second plane can be written in the form \(a_2x + b_2y + c_2z + d_2 = 0\):
- For the second plane \(-3x + 6y - 3z + 2 = 0\):
- \(a_2 = -3\)
- \(b_2 = 6\)
- \(c_2 = -3\)
- \(d_2 = 2\)

### Step 2: Check if the planes are parallel
To check if the planes are parallel, we need to see if the ratios of the coefficients are equal:
\[ \frac{a_1}{a_2} = \frac{1}{-3}, \quad \frac{b_1}{b_2} = \frac{-2}{6} = \frac{-1}{3}, \quad \frac{c_1}{c_2} = \frac{1}{-3} \]
Since all ratios are equal, the planes are parallel.

### Step 3: Calculate the distance between the parallel planes
The formula for the distance \(D\) between two parallel planes \(a_1x + b_1y + c_1z + d_1 = 0\) and \(a_2x + b_2y + c_2z + d_2 = 0\) is given by:
\[ D = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}} \]
Here, we can use the coefficients from either plane since they are parallel. Let's use the coefficients from the second plane:
- \(a = -3\), \(b = 6\), \(c = -3\)
- \(d_1 = -1\) and \(d_2 = 2\)

### Step 4: Substitute values into the distance formula
Substituting the values into the formula:
\[ D = \frac{|2 - (-1)|}{\sqrt{(-3)^2 + 6^2 + (-3)^2}} \]
Calculating the numerator:
\[ |2 + 1| = |3| = 3 \]
Calculating the denominator:
\[ \sqrt{(-3)^2 + 6^2 + (-3)^2} = \sqrt{9 + 36 + 9} = \sqrt{54} \]
Thus, we have:
\[ D = \frac{3}{\sqrt{54}} = \frac{3}{3\sqrt{6}} = \frac{1}{\sqrt{6}} \]

### Final Answer
The distance between the planes is:
\[ D = \frac{1}{\sqrt{6}} \]