If `vec(alpha)=2hati+3hatj-hatk, vec(beta)=-hati+2hatj-4hatk, vecgamma=hati+hatj+hatk`, then `(vec(alpha)xxvec(beta)).(vec(alpha)xxvec(gamma))` is equal to

Options

A. `-74`
B. `74`
C. `64`
D. `60`

Explanation

To solve the problem, we need to find the scalar product of two cross products of vectors \(\vec{\alpha}\), \(\vec{\beta}\), and \(\vec{\gamma}\). The vectors are given as follows:

\[ \vec{\alpha} = 2\hat{i} + 3\hat{j} - \hat{k} \]
\[ \vec{\beta} = -\hat{i} + 2\hat{j} - 4\hat{k} \]
\[ \vec{\gamma} = \hat{i} + \hat{j} + \hat{k} \]

We need to compute \((\vec{\alpha} \times \vec{\beta}) \cdot (\vec{\alpha} \times \vec{\gamma})\).

### Step 1: Calculate \(\vec{\alpha} \times \vec{\beta}\)

To find the cross product \(\vec{\alpha} \times \vec{\beta}\), we set up the determinant:

\[ \vec{\alpha} \times \vec{\beta} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ -1 & 2 & -4 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 3 & -1 \\ 2 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -1 \\ -1 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ -1 & 2 \end{vmatrix} \]

Calculating the minors:

1. For \(\hat{i}\):
\[ = \hat{i} (3 \cdot -4 - (-1) \cdot 2) = \hat{i} (-12 + 2) = -10\hat{i} \]

2. For \(\hat{j}\):
\[ = -\hat{j} (2 \cdot -4 - (-1) \cdot -1) = -\hat{j} (-8 - 1) = 9\hat{j} \]

3. For \(\hat{k}\):
\[ = \hat{k} (2 \cdot 2 - 3 \cdot -1) = \hat{k} (4 + 3) = 7\hat{k} \]

Thus,
\[ \vec{\alpha} \times \vec{\beta} = -10\hat{i} + 9\hat{j} + 7\hat{k} \]

### Step 2: Calculate \(\vec{\alpha} \times \vec{\gamma}\)

Next, we calculate \(\vec{\alpha} \times \vec{\gamma}\):

\[ \vec{\alpha} \times \vec{\gamma} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & 1 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 3 & -1 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} \]

Calculating the minors:

1. For \(\hat{i}\):
\[ = \hat{i} (3 \cdot 1 - (-1) \cdot 1) = \hat{i} (3 + 1) = 4\hat{i} \]

2. For \(\hat{j}\):
\[ = -\hat{j} (2 \cdot 1 - (-1) \cdot 1) = -\hat{j} (2 + 1) = -3\hat{j} \]

3. For \(\hat{k}\):
\[ = \hat{k} (2 \cdot 1 - 3 \cdot 1) = \hat{k} (2 - 3) = -\hat{k} \]

Thus,
\[ \vec{\alpha} \times \vec{\gamma} = 4\hat{i} - 3\hat{j} - \hat{k} \]

### Step 3: Calculate the dot product

Now we compute the dot product:

\[ (\vec{\alpha} \times \vec{\beta}) \cdot (\vec{\alpha} \times \vec{\gamma}) = (-10\hat{i} + 9\hat{j} + 7\hat{k}) \cdot (4\hat{i} - 3\hat{j} - \hat{k}) \]

Calculating the dot product:

\[ = (-10) \cdot 4 + 9 \cdot (-3) + 7 \cdot (-1) \]
\[ = -40 - 27 - 7 \]
\[ = -74 \]

### Final Answer

Thus, the result of \((\vec{\alpha} \times \vec{\beta}) \cdot (\vec{\alpha} \times \vec{\gamma})\) is:

\[ \boxed{-74} \]