Find the distance between the planes 2x - 3y + 6z+8= 0 and `vecr.(2hati-3hatj+6hatk)=-4`

Answer

To find the distance between the two planes given by the equations \(2x - 3y + 6z + 8 = 0\) and \(\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = -4\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the equations of the planes:**
The first plane is already given as:
\[ 2x - 3y + 6z + 8 = 0 \]
We can rewrite it as:
\[ 2x - 3y + 6z = -8 \quad \text{(Plane 1)} \]

The second plane can be rewritten from the vector equation:
\[ \vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = -4 \]
This means:
\[ 2x - 3y + 6z = -4 \quad \text{(Plane 2)} \]

2. **Identify coefficients:**
From both plane equations, we can identify the coefficients:
- For both planes, \(a = 2\), \(b = -3\), \(c = 6\).
- For Plane 1, \(d_1 = -8\).
- For Plane 2, \(d_2 = -4\).

3. **Use the distance formula between parallel planes:**
The formula for the distance \(D\) between two parallel planes of the form \(Ax + By + Cz + d_1 = 0\) and \(Ax + By + Cz + d_2 = 0\) is given by:
\[ D = \frac{|d_1 - d_2|}{\sqrt{A^2 + B^2 + C^2}} \]

4. **Calculate \(d_1 - d_2\):**
\[ d_1 - d_2 = -8 - (-4) = -8 + 4 = -4 \]
Taking the absolute value:
\[ |d_1 - d_2| = 4 \]

5. **Calculate the denominator:**
Now, calculate \(A^2 + B^2 + C^2\):
\[ A^2 + B^2 + C^2 = 2^2 + (-3)^2 + 6^2 = 4 + 9 + 36 = 49 \]
Therefore, \(\sqrt{A^2 + B^2 + C^2} = \sqrt{49} = 7\).

6. **Calculate the distance:**
Substitute the values into the distance formula:
\[ D = \frac{4}{7} \]

### Final Answer:
The distance between the two planes is:
\[ \frac{4}{7} \]