(a) Prove that the normal to the plane containing three points whose position vectors are `vec(a), vec(b), vec(c )` lie in the direction of `vec(b)xx vec(c )+vec(c )xx vec(a)+vec(a)xx vec(b)`.
(b) Find the unit vector perpendicular to the plane ABC, where the position vectors of A, B and C are : `2hat(i)-hat(j)+hat(k), hat(i)+hat(j)+2hat(k)` and `2hat(i)+hat(k)` respectively.

Answer

### Step-by-Step Solution

#### Part (a)

To prove that the normal to the plane containing three points whose position vectors are \(\vec{a}, \vec{b}, \vec{c}\) lies in the direction of \(\vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b}\):

1. **Define the vectors**:
Let \(\vec{A} = \vec{a}\), \(\vec{B} = \vec{b}\), and \(\vec{C} = \vec{c}\).

2. **Find vectors in the plane**:
The vectors lying in the plane can be defined as:
\[ \vec{AB} = \vec{B} - \vec{A} = \vec{b} - \vec{a} \]
\[ \vec{AC} = \vec{C} - \vec{A} = \vec{c} - \vec{a} \]

3. **Normal vector to the plane**:
The normal vector \(\vec{N}\) to the plane formed by \(\vec{AB}\) and \(\vec{AC}\) can be found using the cross product:
\[ \vec{N} = \vec{AB} \times \vec{AC} \]

4. **Express \(\vec{N}\) using cross products**:
Expanding \(\vec{N}\):
\[ \vec{N} = (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) \]
Using properties of cross products:
\[ \vec{N} = \vec{b} \times \vec{c} - \vec{a} \times \vec{c} - \vec{a} \times \vec{b} \]

5. **Rearranging**:
Rearranging the terms gives:
\[ \vec{N} = \vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b} \]

6. **Conclusion**:
Thus, we have shown that the normal to the plane containing the points with position vectors \(\vec{a}, \vec{b}, \vec{c}\) lies in the direction of \(\vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b}\).

#### Part (b)

To find the unit vector perpendicular to the plane ABC, where the position vectors of A, B, and C are given:

1. **Identify position vectors**:
\[ \vec{A} = 2\hat{i} - \hat{j} + \hat{k}, \quad \vec{B} = \hat{i} + \hat{j} + 2\hat{k}, \quad \vec{C} = 2\hat{i} + \hat{k} \]

2. **Calculate vectors AB and AC**:
\[ \vec{AB} = \vec{B} - \vec{A} = (\hat{i} + \hat{j} + 2\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} + 2\hat{j} + \hat{k} \]
\[ \vec{AC} = \vec{C} - \vec{A} = (2\hat{i} + \hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = \hat{j} \]

3. **Find the normal vector using cross product**:
\[ \vec{N} = \vec{AB} \times \vec{AC} \]
Using the determinant method:
\[ \vec{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 0 & 1 & 0 \end{vmatrix} \]

4. **Calculate the determinant**:
\[ \vec{N} = \hat{i}(0 - 1) - \hat{j}(0 - 0) + \hat{k}(-1 \cdot 1 - 0 \cdot 2) = -\hat{i} + 0\hat{j} - \hat{k} = -\hat{i} - \hat{k} \]

5. **Magnitude of the normal vector**:
\[ |\vec{N}| = \sqrt{(-1)^2 + 0^2 + (-1)^2} = \sqrt{2} \]

6. **Unit vector**:
\[ \text{Unit vector} = \frac{\vec{N}}{|\vec{N}|} = \frac{-\hat{i} - \hat{k}}{\sqrt{2}} = -\frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{k} \]