A class XII student appearing for a competitive examination was asked to attempt the following questions.
Let `vec(a),vec(b)` and `vec( c)` be theree non zero vectors.
If `vec(a)=hat(i)-2hat(j),vec(b)=2hat(i)+hat(j)+3hat(k)` then evaluate `(2vec(a)+hat(b))*[(vec(a)+vec(b))xx(vec(a)-2vec(b))]`

Options

A. 0
B. 4
C. 3
D. 2

Answer

To solve the problem, we will evaluate the expression \( (2\vec{a} + \vec{b}) \cdot [(\vec{a} + \vec{b}) \times (\vec{a} - 2\vec{b})] \) step by step.

### Step 1: Define the vectors
Given:
\[ \vec{a} = \hat{i} - 2\hat{j} \]
\[ \vec{b} = 2\hat{i} + \hat{j} + 3\hat{k} \]

### Step 2: Calculate \( 2\vec{a} + \vec{b} \)
First, we calculate \( 2\vec{a} \):
\[ 2\vec{a} = 2(\hat{i} - 2\hat{j}) = 2\hat{i} - 4\hat{j} \]
Now, add \( \vec{b} \):
\[ 2\vec{a} + \vec{b} = (2\hat{i} - 4\hat{j}) + (2\hat{i} + \hat{j} + 3\hat{k}) = (2 + 2)\hat{i} + (-4 + 1)\hat{j} + 3\hat{k} = 4\hat{i} - 3\hat{j} + 3\hat{k} \]

### Step 3: Calculate \( \vec{a} + \vec{b} \)
\[ \vec{a} + \vec{b} = (\hat{i} - 2\hat{j}) + (2\hat{i} + \hat{j} + 3\hat{k}) = (1 + 2)\hat{i} + (-2 + 1)\hat{j} + 3\hat{k} = 3\hat{i} - \hat{j} + 3\hat{k} \]

### Step 4: Calculate \( \vec{a} - 2\vec{b} \)
First, calculate \( 2\vec{b} \):
\[ 2\vec{b} = 2(2\hat{i} + \hat{j} + 3\hat{k}) = 4\hat{i} + 2\hat{j} + 6\hat{k} \]
Now, subtract:
\[ \vec{a} - 2\vec{b} = (\hat{i} - 2\hat{j}) - (4\hat{i} + 2\hat{j} + 6\hat{k}) = (1 - 4)\hat{i} + (-2 - 2)\hat{j} + (-6)\hat{k} = -3\hat{i} - 4\hat{j} - 6\hat{k} \]

### Step 5: Calculate the cross product \( (\vec{a} + \vec{b}) \times (\vec{a} - 2\vec{b}) \)
Using the vectors calculated:
\[ \vec{u} = 3\hat{i} - \hat{j} + 3\hat{k} \]
\[ \vec{v} = -3\hat{i} - 4\hat{j} - 6\hat{k} \]
The cross product is calculated using the determinant:
\[ \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 3 \\ -3 & -4 & -6 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 3 \\ -4 & -6 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 3 \\ -3 & -6 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -1 \\ -3 & -4 \end{vmatrix} \]
Calculating each determinant:
\[ = \hat{i}((-1)(-6) - (3)(-4)) - \hat{j}((3)(-6) - (3)(-3)) + \hat{k}((3)(-4) - (-1)(-3)) \]
\[ = \hat{i}(6 + 12) - \hat{j}(-18 + 9) + \hat{k}(-12 - 3) \]
\[ = 18\hat{i} + 9\hat{j} - 15\hat{k} \]

### Step 6: Calculate the dot product \( (2\vec{a} + \vec{b}) \cdot [(\vec{a} + \vec{b}) \times (\vec{a} - 2\vec{b})] \)
Now we have:
\[ (2\vec{a} + \vec{b}) = 4\hat{i} - 3\hat{j} + 3\hat{k} \]
And:
\[ (\vec{a} + \vec{b}) \times (\vec{a} - 2\vec{b}) = 18\hat{i} + 9\hat{j} - 15\hat{k} \]
Now calculate the dot product:
\[ (4\hat{i} - 3\hat{j} + 3\hat{k}) \cdot (18\hat{i} + 9\hat{j} - 15\hat{k}) = (4)(18) + (-3)(9) + (3)(-15) \]
\[ = 72 - 27 - 45 = 0 \]

### Final Answer
The value of the expression is \( 0 \).

---