Find the vector and cartesian equations of the plane containing the lines :
`vec(r) = hati + 2 hatj - 4 hatk + lambda (2 hati + 3 hatj + 6 hatk)` and
`vec(r) = 3 hati + 3 hatj - 5 hatk + mu (-2 hatj + 3 hatj + 8 hatk)`.

Answer

To find the vector and Cartesian equations of the plane containing the given lines, we will follow these steps:

### Step 1: Identify Points and Direction Ratios
The two lines are given as:
1. Line L1:
\[ \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \]
Here, a point on L1 is \( A(1, 2, -4) \) and the direction vector \( \vec{b_1} = (2, 3, 6) \).

2. Line L2:
\[ \vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(-2\hat{i} + 3\hat{j} + 8\hat{k}) \]
Here, a point on L2 is \( B(3, 3, -5) \) and the direction vector \( \vec{b_2} = (-2, 3, 8) \).

### Step 2: Find the Normal Vector
To find the normal vector \( \vec{n} \) of the plane, we need to compute the cross product of the direction vectors \( \vec{b_1} \) and \( \vec{b_2} \):

\[ \vec{b_1} = \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix}, \quad \vec{b_2} = \begin{pmatrix} -2 \\ 3 \\ 8 \end{pmatrix} \]

The cross product \( \vec{n} = \vec{b_1} \times \vec{b_2} \) can be calculated using the determinant:

\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ -2 & 3 & 8 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{n} = \hat{i}(3 \cdot 8 - 6 \cdot 3) - \hat{j}(2 \cdot 8 - 6 \cdot -2) + \hat{k}(2 \cdot 3 - 3 \cdot -2) \]
\[ = \hat{i}(24 - 18) - \hat{j}(16 + 12) + \hat{k}(6 + 6) \]
\[ = 6\hat{i} - 28\hat{j} + 12\hat{k} \]

Thus, the normal vector is:
\[ \vec{n} = (6, -28, 12) \]

### Step 3: Equation of the Plane
Using the point \( A(1, 2, -4) \) and the normal vector \( \vec{n} = (6, -28, 12) \), the equation of the plane can be written as:
\[ 6(x - 1) - 28(y - 2) + 12(z + 4) = 0 \]

Expanding this:
\[ 6x - 6 - 28y + 56 + 12z + 48 = 0 \]
\[ 6x - 28y + 12z + 98 = 0 \]

### Step 4: Vector Equation of the Plane
The vector equation of the plane can be expressed as:
\[ \vec{r} \cdot \vec{n} = d \]
Where \( d \) is the dot product of the normal vector and a position vector of a point on the plane. Using point \( A(1, 2, -4) \):
\[ d = \vec{n} \cdot \begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix} = 6(1) - 28(2) + 12(-4) = 6 - 56 - 48 = -98 \]
Thus, the vector equation is:
\[ \vec{r} \cdot (6\hat{i} - 28\hat{j} + 12\hat{k}) = -98 \]

### Final Answers
1. **Vector Equation of the Plane:**
\[ \vec{r} \cdot (6\hat{i} - 28\hat{j} + 12\hat{k}) = -98 \]

2. **Cartesian Equation of the Plane:**
\[ 6x - 28y + 12z + 98 = 0 \]