If `veca = hati+hatj-2hatk, vecb=2hati-3hatj+hatk` the value of `(2veca+3vecb) . (2vecbxx3veca)` is :

Options

A. `-4`
B. `16`
C. `10`
D. 0

Explanation

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

### Step 1: Calculate \(2\vec{a} + 3\vec{b}\)

1. First, calculate \(2\vec{a}\):
\[ 2\vec{a} = 2(\hat{i} + \hat{j} - 2\hat{k}) = 2\hat{i} + 2\hat{j} - 4\hat{k} \]

2. Next, calculate \(3\vec{b}\):
\[ 3\vec{b} = 3(2\hat{i} - 3\hat{j} + \hat{k}) = 6\hat{i} - 9\hat{j} + 3\hat{k} \]

3. Now, add \(2\vec{a}\) and \(3\vec{b}\):
\[ 2\vec{a} + 3\vec{b} = (2\hat{i} + 2\hat{j} - 4\hat{k}) + (6\hat{i} - 9\hat{j} + 3\hat{k}) \]
\[ = (2 + 6)\hat{i} + (2 - 9)\hat{j} + (-4 + 3)\hat{k} \]
\[ = 8\hat{i} - 7\hat{j} - \hat{k} \]

### Step 2: Calculate \(2\vec{b} \times 3\vec{a}\)

1. First, calculate \(2\vec{b}\):
\[ 2\vec{b} = 2(2\hat{i} - 3\hat{j} + \hat{k}) = 4\hat{i} - 6\hat{j} + 2\hat{k} \]

2. Next, calculate \(3\vec{a}\):
\[ 3\vec{a} = 3(\hat{i} + \hat{j} - 2\hat{k}) = 3\hat{i} + 3\hat{j} - 6\hat{k} \]

3. Now, calculate the cross product \(2\vec{b} \times 3\vec{a}\):
\[ 2\vec{b} \times 3\vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -6 & 2 \\ 3 & 3 & -6 \end{vmatrix} \]

Expanding the determinant:
\[ = \hat{i}((-6)(-6) - (2)(3)) - \hat{j}((4)(-6) - (2)(3)) + \hat{k}((4)(3) - (-6)(3)) \]
\[ = \hat{i}(36 - 6) - \hat{j}(-24 - 6) + \hat{k}(12 + 18) \]
\[ = 30\hat{i} + 30\hat{j} + 30\hat{k} \]

### Step 3: Calculate the dot product \((2\vec{a} + 3\vec{b}) \cdot (2\vec{b} \times 3\vec{a})\)

1. Now, we take the dot product:
\[ (8\hat{i} - 7\hat{j} - \hat{k}) \cdot (30\hat{i} + 30\hat{j} + 30\hat{k}) \]
\[ = 8 \cdot 30 + (-7) \cdot 30 + (-1) \cdot 30 \]
\[ = 240 - 210 - 30 \]
\[ = 0 \]

Thus, the final answer is:
\[ \boxed{0} \]