The equation of the plane containing the lines `vecr=(hati+hatj-hatk)+lamda(3hati-hatj)` and `vecr=(4hati-hatk)+mu(2hati+3hatk)`, is

Options

A. `vecr.(3hati+9hatj-2hatk)+14=0`
B. `vecr.(3hati+9hatj+2hatk)=14`
C. `vecr.(3hati+9hatj-2hatk)=14`
D. none of these

Explanation

To find the equation 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:
\[ \vec{r} = \hat{i} + \hat{j} - \hat{k} + \lambda(3\hat{i} - \hat{j}) \]
The direction vector \( \vec{b_1} \) of the first line is:
\[ \vec{b_1} = 3\hat{i} - \hat{j} \]

The second line is given by:
\[ \vec{r} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k}) \]
The direction vector \( \vec{b_2} \) of the second line is:
\[ \vec{b_2} = 2\hat{i} + 3\hat{k} \]

### Step 2: Find a point on the plane
We can take a point from either line. Let's take the point from the first line when \( \lambda = 0 \):
\[ \vec{A} = \hat{i} + \hat{j} - \hat{k} = (1, 1, -1) \]

### Step 3: Find the normal vector to the plane
To find the normal vector \( \vec{n} \) to the plane, we take the cross product of the direction vectors \( \vec{b_1} \) and \( \vec{b_2} \):
\[ \vec{n} = \vec{b_1} \times \vec{b_2} = (3\hat{i} - \hat{j}) \times (2\hat{i} + 3\hat{k}) \]

Calculating the cross product:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 0 \\ 2 & 0 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i}((-1)(3) - (0)(0)) - \hat{j}((3)(3) - (0)(2)) + \hat{k}((3)(0) - (-1)(2)) \]
\[ = -3\hat{i} - 9\hat{j} + 2\hat{k} \]
Thus, the normal vector is:
\[ \vec{n} = -3\hat{i} - 9\hat{j} + 2\hat{k} \]

### Step 4: Use the point-normal form of the plane equation
The equation of the plane can be expressed as:
\[ \vec{r} \cdot \vec{n} = \vec{A} \cdot \vec{n} \]
Substituting \( \vec{A} = (1, 1, -1) \) and \( \vec{n} = (-3, -9, 2) \):
\[ \vec{r} \cdot (-3\hat{i} - 9\hat{j} + 2\hat{k}) = (1)(-3) + (1)(-9) + (-1)(2) \]
Calculating the right-hand side:
\[ -3 - 9 - 2 = -14 \]

Thus, the equation becomes:
\[ \vec{r} \cdot (-3\hat{i} - 9\hat{j} + 2\hat{k}) = -14 \]
Rearranging gives:
\[ \vec{r} \cdot (3\hat{i} + 9\hat{j} - 2\hat{k}) = 14 \]

### Final Equation of the Plane
The equation of the plane is:
\[ 3x + 9y - 2z = 14 \]