Find the vector and cartesian equation of the line passing through the point P (1,2,3) and parallel to the planes :
`vec(r).(hati - hatj + 2 hatk) = 5.` `vec(r).(3hati + hatj + hatk) = 6.`

Answer

To find the vector and Cartesian equations of the line passing through the point \( P(1, 2, 3) \) and parallel to the given planes, we can follow these steps:

### Step 1: Identify the normal vectors of the planes

The equations of the planes are given as:
1. \( \vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 5 \)
2. \( \vec{r} \cdot (3\hat{i} + \hat{j} + \hat{k}) = 6 \)

From these equations, we can identify the normal vectors of the planes:
- For Plane 1, the normal vector \( \vec{n_1} = \hat{i} - \hat{j} + 2\hat{k} \)
- For Plane 2, the normal vector \( \vec{n_2} = 3\hat{i} + \hat{j} + \hat{k} \)

### Step 2: Find the direction vector of the line

The line we are looking for is parallel to both planes, which means its direction vector \( \vec{b} \) can be found by taking the cross product of the normal vectors \( \vec{n_1} \) and \( \vec{n_2} \).

\[ \vec{b} = \vec{n_1} \times \vec{n_2} \]

Calculating the cross product:

\[ \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ \vec{b} = \hat{i} \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} \]

Calculating the minors:

\[ \vec{b} = \hat{i}((-1)(1) - (2)(1)) - \hat{j}((1)(1) - (2)(3)) + \hat{k}((1)(1) - (-1)(3)) \]

\[ = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 + 3) \]

\[ = -3\hat{i} + 5\hat{j} + 4\hat{k} \]

Thus, the direction vector of the line is:

\[ \vec{b} = -3\hat{i} + 5\hat{j} + 4\hat{k} \]

### Step 3: Write the vector equation of the line

The vector equation of the line can be expressed as:

\[ \vec{r} = \vec{a} + \lambda \vec{b} \]

where \( \vec{a} \) is the position vector of point \( P(1, 2, 3) \) and \( \lambda \) is a scalar parameter. Therefore,

\[ \vec{r} = (1\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-3\hat{i} + 5\hat{j} + 4\hat{k}) \]

### Step 4: Write the Cartesian equation of the line

To find the Cartesian form, we can express the vector equation in terms of its components:

1. \( x = 1 - 3\lambda \)
2. \( y = 2 + 5\lambda \)
3. \( z = 3 + 4\lambda \)

From these equations, we can eliminate \( \lambda \):

\[ \lambda = \frac{1 - x}{3} = \frac{y - 2}{5} = \frac{z - 3}{4} \]

This gives us the Cartesian equation of the line:

\[ \frac{1 - x}{3} = \frac{y - 2}{5} = \frac{z - 3}{4} \]

### Summary of the Solution

- **Vector Equation of the Line**:
\[ \vec{r} = (1\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-3\hat{i} + 5\hat{j} + 4\hat{k}) \]

- **Cartesian Equation of the Line**:
\[ \frac{1 - x}{3} = \frac{y - 2}{5} = \frac{z - 3}{4} \]