Find the vector equation of the plane passing through three points with position vectors `hati + hatj - 2 hatk, 2 hati - hatj + hatk and hati + 2 hatj + hatk`. Also find the co-ordinates of the point of intersection of this plane and the line `vec(r) = 3 hati - hatj - hatk + lambda (2 hati - 2 hatj + hatk)` .

Answer

To solve the problem, we need to find the vector equation of the plane passing through three points with given position vectors and then find the coordinates of the point of intersection of this plane with a given line.

### Step 1: Identify the Points
Let the position vectors of the three points be:
- \( \vec{A} = \hat{i} + \hat{j} - 2\hat{k} \)
- \( \vec{B} = 2\hat{i} - \hat{j} + \hat{k} \)
- \( \vec{C} = \hat{i} + 2\hat{j} + \hat{k} \)

### Step 2: Find Vectors AB and AC
We need to find the vectors \( \vec{AB} \) and \( \vec{AC} \):
- \( \vec{AB} = \vec{B} - \vec{A} = (2\hat{i} - \hat{j} + \hat{k}) - (\hat{i} + \hat{j} - 2\hat{k}) \)
\[ \vec{AB} = (2 - 1)\hat{i} + (-1 - 1)\hat{j} + (1 + 2)\hat{k} = \hat{i} - 2\hat{j} + 3\hat{k} \]

- \( \vec{AC} = \vec{C} - \vec{A} = (\hat{i} + 2\hat{j} + \hat{k}) - (\hat{i} + \hat{j} - 2\hat{k}) \)
\[ \vec{AC} = (1 - 1)\hat{i} + (2 - 1)\hat{j} + (1 + 2)\hat{k} = 0\hat{i} + \hat{j} + 3\hat{k} \]

### Step 3: Find the Normal Vector
The normal vector \( \vec{n} \) to the plane can be found using the cross product \( \vec{AB} \times \vec{AC} \):
\[ \vec{n} = \vec{AB} \times \vec{AC} = (\hat{i} - 2\hat{j} + 3\hat{k}) \times (0\hat{i} + \hat{j} + 3\hat{k}) \]

Calculating the cross product:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 0 & 1 & 3 \end{vmatrix} \]
\[ = \hat{i} \begin{vmatrix} -2 & 3 \\ 1 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 0 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ 0 & 1 \end{vmatrix} \]
\[ = \hat{i}((-2)(3) - (3)(1)) - \hat{j}((1)(3) - (0)(3)) + \hat{k}((1)(1) - (0)(-2)) \]
\[ = \hat{i}(-6 - 3) - \hat{j}(3) + \hat{k}(1) \]
\[ = -9\hat{i} - 3\hat{j} + 1\hat{k} \]

### Step 4: Equation of the Plane
The equation of the plane can be given by:
\[ \vec{n} \cdot (\vec{r} - \vec{A}) = 0 \]
Substituting \( \vec{n} = -9\hat{i} - 3\hat{j} + \hat{k} \) and \( \vec{A} = \hat{i} + \hat{j} - 2\hat{k} \):
\[ -9(\hat{r} - (\hat{i} + \hat{j} - 2\hat{k})) - 3(\hat{r} - (\hat{i} + \hat{j} - 2\hat{k})) + 1(\hat{r} - (\hat{i} + \hat{j} - 2\hat{k})) = 0 \]
This simplifies to:
\[ -9x - 3y + z + 14 = 0 \]
or
\[ 9x + 3y - z = 14 \]

### Step 5: Find the Intersection with the Line
The line is given by:
\[ \vec{r} = (3\hat{i} - \hat{j} - \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) \]
Let \( \vec{r} = (3 + 2\lambda)\hat{i} + (-1 - 2\lambda)\hat{j} + (-1 + \lambda)\hat{k} \).

Substituting into the plane equation:
\[ 9(3 + 2\lambda) + 3(-1 - 2\lambda) - (-1 + \lambda) = 14 \]
Expanding this:
\[ 27 + 18\lambda - 3 - 6\lambda + 1 - \lambda = 14 \]
\[ 27 - 3 + 1 + (18 - 6 - 1)\lambda = 14 \]
\[ 25 + 11\lambda = 14 \]
\[ 11\lambda = 14 - 25 \]
\[ 11\lambda = -11 \implies \lambda = -1 \]

### Step 6: Find the Coordinates of Intersection Point
Substituting \( \lambda = -1 \) back into the line equation:
\[ \vec{r} = (3 + 2(-1))\hat{i} + (-1 - 2(-1))\hat{j} + (-1 + (-1))\hat{k} \]
\[ = (3 - 2)\hat{i} + (-1 + 2)\hat{j} + (-1 - 1)\hat{k} \]
\[ = 1\hat{i} + 1\hat{j} - 2\hat{k} \]
Thus, the coordinates of the point of intersection are \( (1, 1, -2) \).

### Final Answer
The vector equation of the plane is:
\[ 9x + 3y - z = 14 \]
The coordinates of the point of intersection are:
\[ (1, 1, -2) \]