Find the vector and cartesian forms of the equation of the plane containing two lines :
`vec(r)= (i) + 2 hat(j) - 4 hat(k) + lambda (2 hat(i) + 3 hat(j) + 6 hat(k))`
and ` vec(r) = 3 hat(j) - 5 hat(k) + mu ( - 2 hat(i) + 3 hat(j) + 8 hat(k))`.

Answer

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

### Step 1: Identify Points and Direction Ratios
From the given lines, we can identify:
- For Line L1:
- Point \( A(1, 2, -4) \)
- Direction ratios \( \vec{b_1} = (2, 3, 6) \)

- For Line L2:
- Point \( B(0, 3, -5) \)
- Direction ratios \( \vec{b_2} = (-2, 3, 8) \)

### Step 2: Find the Normal Vector
The normal vector \( \vec{n} \) to the plane can be found using the cross product of the direction ratios of the two lines:
\[ \vec{n} = \vec{b_1} \times \vec{b_2} \]

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

Calculating the determinant:
\[ \vec{n} = \hat{i} \begin{vmatrix} 3 & 6 \\ 3 & 8 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 6 \\ -2 & 8 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ -2 & 3 \end{vmatrix} \]
\[ = \hat{i} (3 \cdot 8 - 6 \cdot 3) - \hat{j} (2 \cdot 8 - 6 \cdot -2) + \hat{k} (2 \cdot 3 - 3 \cdot -2) \]
\[ = \hat{i} (24 - 18) - \hat{j} (16 + 12) + \hat{k} (6 + 6) \]
\[ = 6\hat{i} - 28\hat{j} + 12\hat{k} \]

Thus, the normal vector \( \vec{n} = (6, -28, 12) \).

### Step 3: Equation of the Plane in Vector Form
The equation of the plane can be expressed in vector form as:
\[ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \]
Where \( \vec{a} \) is a position vector of a point on the plane (we can use point \( A \)):
\[ \vec{a} = (1, 2, -4) \]
Calculating \( \vec{a} \cdot \vec{n} \):
\[ \vec{a} \cdot \vec{n} = (1, 2, -4) \cdot (6, -28, 12) = 6 \cdot 1 - 28 \cdot 2 + 12 \cdot (-4) \]
\[ = 6 - 56 - 48 = -98 \]

Thus, the vector form of the equation of the plane is:
\[ \vec{r} \cdot (6\hat{i} - 28\hat{j} + 12\hat{k}) = -98 \]

### Step 4: Convert to Cartesian Form
To convert to Cartesian form, we express \( \vec{r} = (x, y, z) \):
\[ 6x - 28y + 12z = -98 \]
Rearranging gives:
\[ 6x - 28y + 12z + 98 = 0 \]

### Final Answers
- **Vector Form**: \( \vec{r} \cdot (6\hat{i} - 28\hat{j} + 12\hat{k}) = -98 \)
- **Cartesian Form**: \( 6x - 28y + 12z + 98 = 0 \)