Find the vector and cartesian equation of the plane :
(i) that passes through the point (5,2,-4) and perpendicular to the line with direction-ratios `lt 2, 3, -1 gt `
(ii) that passes through the point (1,0, -2) and the normal to the plane is `hati + hatj - hatk `
(iii) that passes through the point (1,4,6) and the normal vector to the plane is `hati - 2 hatj + hatk . `

Answer

To find the vector and Cartesian equations of the planes as described in the problem, we will follow a systematic approach for each part.

### (i) Plane through the point (5,2,-4) and perpendicular to the line with direction ratios \( \langle 2, 3, -1 \rangle \)

1. **Identify the point and direction ratios**:
- Point \( P(5, 2, -4) \)
- Direction ratios (normal vector) \( \mathbf{n} = \langle 2, 3, -1 \rangle \)

2. **Vector equation of the plane**:
The vector equation of a plane can be written as:
\[ \mathbf{r} \cdot \mathbf{n} = d \]
where \( \mathbf{r} = \langle x, y, z \rangle \) and \( d = \mathbf{n} \cdot \mathbf{p} \) (where \( \mathbf{p} \) is the position vector of point \( P \)).
\[ d = \langle 2, 3, -1 \rangle \cdot \langle 5, 2, -4 \rangle = 2 \cdot 5 + 3 \cdot 2 + (-1) \cdot (-4) = 10 + 6 + 4 = 20 \]
Thus, the vector equation is:
\[ \mathbf{r} \cdot \langle 2, 3, -1 \rangle = 20 \]

3. **Cartesian equation of the plane**:
The Cartesian form can be derived from the normal vector and the point:
\[ 2(x - 5) + 3(y - 2) - 1(z + 4) = 0 \]
Simplifying this:
\[ 2x - 10 + 3y - 6 - z - 4 = 0 \implies 2x + 3y - z - 20 = 0 \]

### (ii) Plane through the point (1,0,-2) with normal vector \( \langle 1, 1, -1 \rangle \)

1. **Identify the point and normal vector**:
- Point \( P(1, 0, -2) \)
- Normal vector \( \mathbf{n} = \langle 1, 1, -1 \rangle \)

2. **Vector equation of the plane**:
\[ d = \mathbf{n} \cdot \mathbf{p} = \langle 1, 1, -1 \rangle \cdot \langle 1, 0, -2 \rangle = 1 \cdot 1 + 1 \cdot 0 + (-1) \cdot (-2) = 1 + 0 + 2 = 3 \]
Thus, the vector equation is:
\[ \mathbf{r} \cdot \langle 1, 1, -1 \rangle = 3 \]

3. **Cartesian equation of the plane**:
\[ 1(x - 1) + 1(y - 0) - 1(z + 2) = 0 \]
Simplifying this:
\[ x - 1 + y - z - 2 = 0 \implies x + y - z - 3 = 0 \]

### (iii) Plane through the point (1,4,6) with normal vector \( \langle 1, -2, 1 \rangle \)

1. **Identify the point and normal vector**:
- Point \( P(1, 4, 6) \)
- Normal vector \( \mathbf{n} = \langle 1, -2, 1 \rangle \)

2. **Vector equation of the plane**:
\[ d = \mathbf{n} \cdot \mathbf{p} = \langle 1, -2, 1 \rangle \cdot \langle 1, 4, 6 \rangle = 1 \cdot 1 + (-2) \cdot 4 + 1 \cdot 6 = 1 - 8 + 6 = -1 \]
Thus, the vector equation is:
\[ \mathbf{r} \cdot \langle 1, -2, 1 \rangle = -1 \]

3. **Cartesian equation of the plane**:
\[ 1(x - 1) - 2(y - 4) + 1(z - 6) = 0 \]
Simplifying this:
\[ x - 1 - 2y + 8 + z - 6 = 0 \implies x - 2y + z + 1 = 0 \]

### Summary of Results:
1. **Part (i)**:
- Vector equation: \( \mathbf{r} \cdot \langle 2, 3, -1 \rangle = 20 \)
- Cartesian equation: \( 2x + 3y - z - 20 = 0 \)

2. **Part (ii)**:
- Vector equation: \( \mathbf{r} \cdot \langle 1, 1, -1 \rangle = 3 \)
- Cartesian equation: \( x + y - z - 3 = 0 \)

3. **Part (iii)**:
- Vector equation: \( \mathbf{r} \cdot \langle 1, -2, 1 \rangle = -1 \)
- Cartesian equation: \( x - 2y + z + 1 = 0 \)