(i) Find the vector equation of the plane through the intersection of the planes :
`vec(r) . (hati + hatj + hatk) = 6, vec(r) . (2 hati + 3 hatj + 4 hatk) = -5 `
and the point (1,1,1).
(ii) Find the equation of the plane which contains the line of intersection of the planes :
`vec(r) . (hat(i) + 2 hat(j) + 3 hat(k) ) - 4 = 0. vec(r). (2 hat(i) + hatj - hat(k) ) + 5 = 0`
and which is perpendicular to the plane :
`vec(r) . (5 hati + 3 hatj - 6 hatk )` + 8 = 0 .
(iii) Find the equation the plane passing through the intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0 and the point (1,1,1) .

Answer

To solve the given questions step by step, we will derive the vector equations of the planes as required.

### (i) Find the vector equation of the plane through the intersection of the planes:
1. **Identify the equations of the given planes**:
- Plane 1: \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 6 \)
- Plane 2: \( \vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5 \)

2. **Convert the equations into standard form**:
- Plane 1: \( \vec{r} \cdot \vec{n_1} = d_1 \) where \( \vec{n_1} = \hat{i} + \hat{j} + \hat{k} \) and \( d_1 = 6 \).
- Plane 2: \( \vec{r} \cdot \vec{n_2} = d_2 \) where \( \vec{n_2} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and \( d_2 = -5 \).

3. **Write the general equation of the plane through the intersection**:
The equation of the plane through the intersection of these two planes can be expressed as:
\[ \vec{r} \cdot \vec{n_1} + \lambda (\vec{r} \cdot \vec{n_2}) = d_1 + \lambda d_2 \]
where \( \lambda \) is a scalar parameter.

4. **Substituting the known values**:
\[ \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) + \lambda (\vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k})) = 6 + \lambda(-5) \]

5. **Using the point (1,1,1)**:
Substitute \( \vec{r} = (1,1,1) \):
\[ (1 + 1 + 1) + \lambda (2 + 3 + 4) = 6 - 5\lambda \]
\[ 3 + 9\lambda = 6 - 5\lambda \]
\[ 14\lambda = 3 \implies \lambda = \frac{3}{14} \]

6. **Final equation of the plane**:
Substitute \( \lambda \) back into the equation:
\[ \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) + \frac{3}{14} (\vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k})) = 6 - \frac{15}{14} \]
Simplifying gives:
\[ \vec{r} \cdot \left(1 + \frac{6}{14}\hat{i} + \frac{9}{14}\hat{j} + \frac{12}{14}\hat{k}\right) = \frac{69}{14} \]