Find the vector and catesian equations of the plane containing the lines :
`vec(r) = 2 hati + hatj - 3 hatk + lambda (hati + 2 hatj + 5 hatk )`
and `vec(r) = 3 hati + 3 hatj - 7 hatk + mu (3 hati - 2 hatj + 5 hatk )`.

Answer

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

### Step 1: Identify the Direction Vectors of the Lines
The first line is given by:
\[ \vec{r_1} = 2\hat{i} + \hat{j} - 3\hat{k} + \lambda(\hat{i} + 2\hat{j} + 5\hat{k}) \]
The direction vector of the first line, denoted as \(\vec{A}\), is:
\[ \vec{A} = \hat{i} + 2\hat{j} + 5\hat{k} \]

The second line is given by:
\[ \vec{r_2} = 3\hat{i} + 3\hat{j} - 7\hat{k} + \mu(3\hat{i} - 2\hat{j} + 5\hat{k}) \]
The direction vector of the second line, denoted as \(\vec{B}\), is:
\[ \vec{B} = 3\hat{i} - 2\hat{j} + 5\hat{k} \]

### Step 2: Find the Normal Vector to the Plane
The normal vector \(\vec{N}\) to the plane can be found by taking the cross product of the direction vectors \(\vec{A}\) and \(\vec{B}\):
\[ \vec{N} = \vec{A} \times \vec{B} \]
Calculating the cross product:
\[ \vec{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 5 \\ 3 & -2 & 5 \end{vmatrix} \]
Calculating the determinant:
\[ \vec{N} = \hat{i}(2 \cdot 5 - 5 \cdot (-2)) - \hat{j}(1 \cdot 5 - 5 \cdot 3) + \hat{k}(1 \cdot (-2) - 2 \cdot 3) \]
\[ = \hat{i}(10 + 10) - \hat{j}(5 - 15) + \hat{k}(-2 - 6) \]
\[ = 20\hat{i} + 10\hat{j} - 8\hat{k} \]

### Step 3: Use a Point on the Plane
We can use a point from either line to find the equation of the plane. Let's use the point from the first line:
\[ \vec{P} = 2\hat{i} + \hat{j} - 3\hat{k} \]

### Step 4: Write the Vector Equation of the Plane
The vector equation of the plane can be expressed as:
\[ \vec{r} \cdot \vec{N} = \vec{P} \cdot \vec{N} \]
Calculating \(\vec{P} \cdot \vec{N}\):
\[ \vec{P} \cdot \vec{N} = (2\hat{i} + \hat{j} - 3\hat{k}) \cdot (20\hat{i} + 10\hat{j} - 8\hat{k}) \]
\[ = 2 \cdot 20 + 1 \cdot 10 - 3 \cdot (-8) = 40 + 10 + 24 = 74 \]
Thus, the equation becomes:
\[ \vec{r} \cdot (20\hat{i} + 10\hat{j} - 8\hat{k}) = 74 \]

### Step 5: Write the Cartesian Equation of the Plane
Let \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\). Then, the equation becomes:
\[ 20x + 10y - 8z = 74 \]
To convert it to standard form, we can divide the entire equation by 2:
\[ 10x + 5y - 4z = 37 \]

### Final Answers
- **Vector Equation of the Plane**: \(\vec{r} \cdot (20\hat{i} + 10\hat{j} - 8\hat{k}) = 74\)
- **Cartesian Equation of the Plane**: \(10x + 5y - 4z = 37\)