Convert the equation of the plane `vecr = (hati-hatj)+lambda(-hati+hatj+2hatk)+mu(hati+2hatj+hatk)` into scalar product form.

Answer

To convert the given vector equation of the plane into scalar product form, we will follow these steps:

### Step 1: Identify the vectors
Given the equation of the plane:
\[ \vec{r} = \hat{i} - \hat{j} + \lambda(-\hat{i} + \hat{j} + 2\hat{k}) + \mu(\hat{i} + 2\hat{j} + \hat{k}) \]
We can identify the following vectors:
- Position vector \( \vec{A} = \hat{i} - \hat{j} \)
- Direction vector \( \vec{B} = -\hat{i} + \hat{j} + 2\hat{k} \)
- Direction vector \( \vec{C} = \hat{i} + 2\hat{j} + \hat{k} \)

### Step 2: Find the normal vector
The normal vector \( \vec{n} \) to the plane can be found using the cross product of the direction vectors \( \vec{B} \) and \( \vec{C} \):
\[ \vec{B} = \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}, \quad \vec{C} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \]
We calculate \( \vec{n} = \vec{B} \times \vec{C} \) using the determinant method:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ 1 & 2 & 1 \end{vmatrix} \]

### Step 3: Expand the determinant
Calculating the determinant:
\[ \vec{n} = \hat{i} \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = 1 \cdot 1 - 2 \cdot 2 = 1 - 4 = -3 \)
2. \( \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} = -1 \cdot 1 - 2 \cdot 1 = -1 - 2 = -3 \)
3. \( \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} = -1 \cdot 2 - 1 \cdot 1 = -2 - 1 = -3 \)

Thus:
\[ \vec{n} = -3\hat{i} + 3\hat{j} - 3\hat{k} = -3(\hat{i} - \hat{j} + \hat{k}) \]

### Step 4: Write the equation of the plane
The equation of the plane in scalar product form is given by:
\[ \vec{r} - \vec{A} \cdot \vec{n} = 0 \]
Substituting \( \vec{A} \) and \( \vec{n} \):
\[ \vec{r} \cdot \vec{n} = \vec{A} \cdot \vec{n} \]

### Step 5: Calculate \( \vec{A} \cdot \vec{n} \)
Calculating \( \vec{A} \cdot \vec{n} \):
\[ \vec{A} = \hat{i} - \hat{j}, \quad \vec{n} = -3(\hat{i} - \hat{j} + \hat{k}) \]
\[ \vec{A} \cdot \vec{n} = (\hat{i} - \hat{j}) \cdot (-3(\hat{i} - \hat{j} + \hat{k})) = -3(\hat{i} \cdot \hat{i} - \hat{i} \cdot \hat{j} - \hat{j} \cdot \hat{i} + \hat{j} \cdot \hat{j}) \]
This simplifies to:
\[ -3(1 + 1) = -3 \cdot 2 = -6 \]

### Final Equation
Thus, the scalar product form of the equation of the plane is:
\[ \vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 2 \]