Find the vector equation of a plane which is at a distance of 7 units from the origin and which is normal to the vector `3 hati + 5 hatj - 6 hatk`.
(ii) Find the vector equation of a plane , which is at a distance of 5 units from the origin and its normal vector is
2 ` hati - 3 hatj + 6 hatk `.

Answer

To find the vector equation of the planes as described in the question, we will follow these steps:

### Part (i)

1. **Identify the normal vector and distance from the origin**:
- The normal vector is given as \( \mathbf{n} = 3 \hat{i} + 5 \hat{j} - 6 \hat{k} \).
- The distance from the origin is \( p = 7 \) units.

2. **Calculate the magnitude of the normal vector**:
\[ |\mathbf{n}| = \sqrt{3^2 + 5^2 + (-6)^2} = \sqrt{9 + 25 + 36} = \sqrt{70} \]

3. **Find the unit normal vector**:
\[ \hat{n} = \frac{\mathbf{n}}{|\mathbf{n}|} = \frac{3 \hat{i} + 5 \hat{j} - 6 \hat{k}}{\sqrt{70}} = \frac{3}{\sqrt{70}} \hat{i} + \frac{5}{\sqrt{70}} \hat{j} - \frac{6}{\sqrt{70}} \hat{k} \]

4. **Use the formula for the equation of the plane**:
The vector equation of the plane can be expressed as:
\[ \mathbf{r} \cdot \hat{n} = p \]
Substituting the values we have:
\[ \mathbf{r} \cdot \left( \frac{3}{\sqrt{70}} \hat{i} + \frac{5}{\sqrt{70}} \hat{j} - \frac{6}{\sqrt{70}} \hat{k} \right) = 7 \]

5. **Rewrite the equation**:
\[ \mathbf{r} \cdot \left( \frac{3}{\sqrt{70}} \hat{i} + \frac{5}{\sqrt{70}} \hat{j} - \frac{6}{\sqrt{70}} \hat{k} \right) - 7 = 0 \]

### Part (ii)

1. **Identify the normal vector and distance from the origin**:
- The normal vector is given as \( \mathbf{n} = 2 \hat{i} - 3 \hat{j} + 6 \hat{k} \).
- The distance from the origin is \( p = 5 \) units.

2. **Calculate the magnitude of the normal vector**:
\[ |\mathbf{n}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

3. **Find the unit normal vector**:
\[ \hat{n} = \frac{\mathbf{n}}{|\mathbf{n}|} = \frac{2 \hat{i} - 3 \hat{j} + 6 \hat{k}}{7} = \frac{2}{7} \hat{i} - \frac{3}{7} \hat{j} + \frac{6}{7} \hat{k} \]

4. **Use the formula for the equation of the plane**:
\[ \mathbf{r} \cdot \hat{n} = p \]
Substituting the values we have:
\[ \mathbf{r} \cdot \left( \frac{2}{7} \hat{i} - \frac{3}{7} \hat{j} + \frac{6}{7} \hat{k} \right) = 5 \]

5. **Rewrite the equation**:
\[ \mathbf{r} \cdot \left( \frac{2}{7} \hat{i} - \frac{3}{7} \hat{j} + \frac{6}{7} \hat{k} \right) - 5 = 0 \]

### Summary of the Vector Equations:
1. For the first plane:
\[ \mathbf{r} \cdot \left( \frac{3}{\sqrt{70}} \hat{i} + \frac{5}{\sqrt{70}} \hat{j} - \frac{6}{\sqrt{70}} \hat{k} \right) - 7 = 0 \]

2. For the second plane:
\[ \mathbf{r} \cdot \left( \frac{2}{7} \hat{i} - \frac{3}{7} \hat{j} + \frac{6}{7} \hat{k} \right) - 5 = 0 \]