Find the value of `alphatimes(betatimesgamma)`, where `alpha=2hat(i)-10hat(j)+2hat(k), beta=3hat(i)+hat(j)+2hat(k), gamma=2hat(i)+hat(j)+3hat(k)`.

Answer

To find the value of \( \alpha \times (\beta \times \gamma) \), we will follow these steps:

### Step 1: Define the vectors
Given:
\[ \alpha = 2\hat{i} - 10\hat{j} + 2\hat{k} \]
\[ \beta = 3\hat{i} + \hat{j} + 2\hat{k} \]
\[ \gamma = 2\hat{i} + \hat{j} + 3\hat{k} \]

### Step 2: Calculate \( \beta \times \gamma \)
To find \( \beta \times \gamma \), we will use the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of \( \beta \) and \( \gamma \):

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

### Step 3: Calculate the determinant
Expanding the determinant, we have:

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

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = (1)(3) - (1)(2) = 3 - 2 = 1 \)
2. \( \begin{vmatrix} 3 & 2 \\ 2 & 3 \end{vmatrix} = (3)(3) - (2)(2) = 9 - 4 = 5 \)
3. \( \begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} = (3)(1) - (1)(2) = 3 - 2 = 1 \)

Putting it all together:
\[ \beta \times \gamma = \hat{i}(1) - \hat{j}(5) + \hat{k}(1) = \hat{i} - 5\hat{j} + \hat{k} \]

### Step 4: Calculate \( \alpha \times (\beta \times \gamma) \)
Now we need to calculate \( \alpha \times (\beta \times \gamma) \):
\[ \alpha \times (\beta \times \gamma) = \alpha \times (\hat{i} - 5\hat{j} + \hat{k}) \]

Using the determinant:
\[ \alpha \times (\beta \times \gamma) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -10 & 2 \\ 1 & -5 & 1 \end{vmatrix} \]

### Step 5: Calculate the determinant
Expanding this determinant:
\[ \alpha \times (\beta \times \gamma) = \hat{i} \begin{vmatrix} -10 & 2 \\ -5 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 2 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -10 \\ 1 & -5 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -10 & 2 \\ -5 & 1 \end{vmatrix} = (-10)(1) - (2)(-5) = -10 + 10 = 0 \)
2. \( \begin{vmatrix} 2 & 2 \\ 1 & 1 \end{vmatrix} = (2)(1) - (2)(1) = 2 - 2 = 0 \)
3. \( \begin{vmatrix} 2 & -10 \\ 1 & -5 \end{vmatrix} = (2)(-5) - (-10)(1) = -10 + 10 = 0 \)

Putting it all together:
\[ \alpha \times (\beta \times \gamma) = \hat{i}(0) - \hat{j}(0) + \hat{k}(0) = 0 \]

### Final Answer
Thus, the value of \( \alpha \times (\beta \times \gamma) \) is the zero vector:
\[ \alpha \times (\beta \times \gamma) = \mathbf{0} \]