The vector equation of the plane passing through the points `hati+hatj-2hatk,2hati-hatj+hatk` and `hati+hatj+hatk`, is

Options

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

Explanation

To find the vector equation of the plane passing through the points given in vector form, we can follow these steps:

### Step 1: Identify the Points
We have three points in vector form:
- \( A = \hat{i} + \hat{j} - 2\hat{k} \)
- \( B = 2\hat{i} - \hat{j} + \hat{k} \)
- \( C = \hat{i} + \hat{j} + \hat{k} \)

### Step 2: Find Vectors in the Plane
We can find two vectors that lie in the plane by subtracting the position vectors of these points:
- \( \vec{AB} = \vec{B} - \vec{A} = (2\hat{i} - \hat{j} + \hat{k}) - (\hat{i} + \hat{j} - 2\hat{k}) \)
\[ \vec{AB} = (2 - 1)\hat{i} + (-1 - 1)\hat{j} + (1 + 2)\hat{k} = \hat{i} - 2\hat{j} + 3\hat{k} \]

- \( \vec{BC} = \vec{C} - \vec{B} = (\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) \)
\[ \vec{BC} = (1 - 2)\hat{i} + (1 + 1)\hat{j} + (1 - 1)\hat{k} = -\hat{i} + 2\hat{j} + 0\hat{k} \]

### Step 3: Find the Normal Vector
To find the normal vector \( \vec{n} \) to the plane, we take the cross product of \( \vec{AB} \) and \( \vec{BC} \):
\[ \vec{n} = \vec{AB} \times \vec{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ -1 & 2 & 0 \end{vmatrix} \]

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

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -2 & 3 \\ 2 & 0 \end{vmatrix} = (-2)(0) - (3)(2) = -6 \)
2. \( \begin{vmatrix} 1 & 3 \\ -1 & 0 \end{vmatrix} = (1)(0) - (3)(-1) = 3 \)
3. \( \begin{vmatrix} 1 & -2 \\ -1 & 2 \end{vmatrix} = (1)(2) - (-2)(-1) = 2 - 2 = 0 \)

Putting it all together:
\[ \vec{n} = -6\hat{i} - 3\hat{j} + 0\hat{k} = -6\hat{i} - 3\hat{j} \]

### Step 4: Simplify the Normal Vector
We can factor out -3:
\[ \vec{n} = -3(2\hat{i} + \hat{j}) \]

### Step 5: Use a Point to Write the Plane Equation
Using point \( C \) (or any point) to write the equation of the plane:
\[ \vec{r} \cdot \vec{n} = \vec{C} \cdot \vec{n} \]
Where \( \vec{r} \) is any point on the plane.

Calculating \( \vec{C} \cdot \vec{n} \):
\[ \vec{C} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{n} = -3(2\hat{i} + \hat{j}) \]
\[ \vec{C} \cdot \vec{n} = (1)(-6) + (1)(-3) + (0)(0) = -6 - 3 = -9 \]

Thus, the equation of the plane can be written as:
\[ \vec{r} \cdot (2\hat{i} + \hat{j}) = -3 \]

### Final Answer
The vector equation of the plane is:
\[ \vec{r} \cdot (2\hat{i} + \hat{j}) = 3 \]