If `veca=2hati + hatj+ hatk, vecb= hati+ 2hatj + 2hatk,vecc = hati+ hatj + 2hatk and (1 + alpha) hati+ beta(1+ alpha)hatj+gamma(1+alpha)(1 + beta) hatk= vecaxx(vecb xx vecc) , " then " alpha, beta and gamma " are "`

Options

A. `-2,-4,-2/3`
B. `2,-4,2/3`
C. `-2,4,2/3`
D. `2,4,-2/3`

Explanation

To solve the given problem step by step, we will follow the vector algebra rules and properties.

### Step 1: Define the vectors
We are given the vectors:
- \(\vec{a} = 2\hat{i} + \hat{j} + \hat{k}\)
- \(\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}\)
- \(\vec{c} = \hat{i} + \hat{j} + 2\hat{k}\)

### Step 2: Compute \(\vec{b} \times \vec{c}\)
We will first find the cross product \(\vec{b} \times \vec{c}\).

\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 1 & 1 & 2 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} \]

Calculating the individual determinants:

\[ = \hat{i} (2 \cdot 2 - 2 \cdot 1) - \hat{j} (1 \cdot 2 - 2 \cdot 1) + \hat{k} (1 \cdot 1 - 2 \cdot 1) \]
\[ = \hat{i} (4 - 2) - \hat{j} (2 - 2) + \hat{k} (1 - 2) \]
\[ = 2\hat{i} + 0\hat{j} - \hat{k} = 2\hat{i} - \hat{k} \]

### Step 3: Compute \(\vec{a} \times (\vec{b} \times \vec{c})\)
Now we need to compute \(\vec{a} \times (\vec{b} \times \vec{c})\):

\[ \vec{a} \times (2\hat{i} - \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 2 & 0 & -1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 0 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 2 & 0 \end{vmatrix} \]

Calculating the individual determinants:

\[ = \hat{i} (1 \cdot -1 - 1 \cdot 0) - \hat{j} (2 \cdot -1 - 1 \cdot 2) + \hat{k} (2 \cdot 0 - 1 \cdot 2) \]
\[ = -\hat{i} - \hat{j} (-2 - 2) - 2\hat{k} \]
\[ = -\hat{i} + 4\hat{j} - 2\hat{k} \]

### Step 4: Set the equation
We are given that:

\[ (1 + \alpha)\hat{i} + \beta(1 + \alpha)\hat{j} + \gamma(1 + \alpha)(1 + \beta)\hat{k} = -\hat{i} + 4\hat{j} - 2\hat{k} \]

### Step 5: Compare coefficients
Now, we can compare the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\):

1. For \(\hat{i}\):
\[ 1 + \alpha = -1 \implies \alpha = -2 \]

2. For \(\hat{j}\):
\[ \beta(1 + \alpha) = 4 \implies \beta(-1) = 4 \implies \beta = -4 \]

3. For \(\hat{k}\):
\[ \gamma(1 + \alpha)(1 + \beta) = -2 \]
Substituting \(\alpha = -2\) and \(\beta = -4\):
\[ \gamma(-1)(-3) = -2 \implies 3\gamma = -2 \implies \gamma = -\frac{2}{3} \]

### Final Values
Thus, the values of \(\alpha\), \(\beta\), and \(\gamma\) are:
- \(\alpha = -2\)
- \(\beta = -4\)
- \(\gamma = -\frac{2}{3}\)