A class XII student appearing for a competitive examination was asked to attempt the following questions
Let `veca,vecb` and `vecc` be three non zero vectors.
If `veca=hati-2hatj,vecb=2hati+hatj+3hatk` then evaluate `(2veca+vecb).[(veca+vecb)xx(veca-2vecb)]`

Options

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

Answer

To solve the problem, we need to evaluate the expression \((2\vec{a} + \vec{b}) \cdot ((\vec{a} + \vec{b}) \times (\vec{a} - 2\vec{b}))\) given the vectors \(\vec{a} = \hat{i} - 2\hat{j}\) and \(\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}\).

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

### Step 2: 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 3: Calculate \(\vec{a} - 2\vec{b}\)
Calculating \(2\vec{b}\):
\[ 2\vec{b} = 2(2\hat{i} + \hat{j} + 3\hat{k}) = 4\hat{i} + 2\hat{j} + 6\hat{k} \]
Now, subtracting from \(\vec{a}\):
\[ \vec{a} - 2\vec{b} = (\hat{i} - 2\hat{j}) - (4\hat{i} + 2\hat{j} + 6\hat{k}) = (1 - 4)\hat{i} + (-2 - 2)\hat{j} + (0 - 6)\hat{k} = -3\hat{i} - 4\hat{j} - 6\hat{k} \]

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

Putting it all together:
\[ \vec{u} \times \vec{v} = 18\hat{i} + 9\hat{j} - 15\hat{k} = 18\hat{i} + 9\hat{j} - 15\hat{k} \]

### Step 5: Calculate \((2\vec{a} + \vec{b}) \cdot ((\vec{a} + \vec{b}) \times (\vec{a} - 2\vec{b}))\)
Now we need to calculate the dot product:
\[ (4\hat{i} - 3\hat{j} + 3\hat{k}) \cdot (18\hat{i} + 9\hat{j} - 15\hat{k}) = 4 \cdot 18 + (-3) \cdot 9 + 3 \cdot (-15) \]
Calculating:
\[ = 72 - 27 - 45 = 72 - 72 = 0 \]

### Final Answer
Thus, the value of \((2\vec{a} + \vec{b}) \cdot ((\vec{a} + \vec{b}) \times (\vec{a} - 2\vec{b}))\) is \(\boxed{0}\).