Find the vector and cartesan equation of the plane passing through the poin (1,2,-4) and parallel to the lines.
`vec(r ) = hat(i) + 2 hat(j) - 4 hat(k) + lambda (2 hat(i) + 3 hat(j) + 6 hat(k))`
and `vec(r) = hat(i) - 3 hat(j) + 5 hat(k) + mu (hat(i) + hat(j) - hat(k))` .

Answer

To find the vector and Cartesian equations of the plane passing through the point \( P(1, 2, -4) \) and parallel to the given lines, we can follow these steps:

### Step 1: Identify Direction Vectors
The direction vectors of the lines can be extracted from their vector equations.

For the first line:
\[ \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \]
The direction vector \( \vec{b_1} \) is:
\[ \vec{b_1} = 2\hat{i} + 3\hat{j} + 6\hat{k} \]

For the second line:
\[ \vec{r} = \hat{i} - 3\hat{j} + 5\hat{k} + \mu(\hat{i} + \hat{j} - \hat{k}) \]
The direction vector \( \vec{b_2} \) is:
\[ \vec{b_2} = \hat{i} + \hat{j} - \hat{k} \]

### Step 2: Find the Normal Vector
The normal vector \( \vec{n} \) of the plane can be found by taking the cross product of the two direction vectors \( \vec{b_1} \) and \( \vec{b_2} \).

\[ \vec{n} = \vec{b_1} \times \vec{b_2} \]

Calculating the cross product:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i} \begin{vmatrix} 3 & 6 \\ 1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 6 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
\[ = \hat{i}(3 \cdot -1 - 6 \cdot 1) - \hat{j}(2 \cdot -1 - 6 \cdot 1) + \hat{k}(2 \cdot 1 - 3 \cdot 1) \]
\[ = \hat{i}(-3 - 6) - \hat{j}(-2 - 6) + \hat{k}(2 - 3) \]
\[ = -9\hat{i} + 8\hat{j} - 1\hat{k} \]

Thus, the normal vector is:
\[ \vec{n} = -9\hat{i} + 8\hat{j} - \hat{k} \]

### Step 3: Write the Vector Equation of the Plane
The vector equation of the plane can be expressed as:
\[ \vec{r} \cdot \vec{n} = \vec{P} \cdot \vec{n} \]
Where \( \vec{P} = (1, 2, -4) \).

Calculating \( \vec{P} \cdot \vec{n} \):
\[ \vec{P} \cdot \vec{n} = 1(-9) + 2(8) + (-4)(-1) = -9 + 16 + 4 = 11 \]

Thus, the vector equation of the plane is:
\[ \vec{r} \cdot (-9\hat{i} + 8\hat{j} - \hat{k}) = 11 \]

### Step 4: Write the Cartesian Equation of the Plane
The Cartesian equation can be derived from the vector equation:
\[ -9x + 8y - z = 11 \]

Rearranging gives:
\[ 9x - 8y + z = 11 \]

### Final Answers
- **Vector Equation of the Plane**:
\[ \vec{r} \cdot (-9\hat{i} + 8\hat{j} - \hat{k}) = 11 \]
- **Cartesian Equation of the Plane**:
\[ 9x - 8y + z = 11 \]