Let ` vec(a) = hat(i) - alpha hat(j) + beta hat(k) , vec(b) = 3 hat(i) + beta hat(j) - alpha hat(k) and vec(c) = - alpha hat(i) - 2 hat(j) + hat(k)` , where ` alpha and beta ` are integers. If ` vec(a)* vec(b) = - 1 and vec(b) * vec(c) = 10 " then " ( vec(a) xx vec(b)) * vec(c)` is equal to _____

Answer

To solve the problem step by step, we will follow the given conditions and perform the necessary calculations.

### Step 1: Define the vectors
Given:
- \( \vec{a} = \hat{i} - \alpha \hat{j} + \beta \hat{k} \)
- \( \vec{b} = 3 \hat{i} + \beta \hat{j} - \alpha \hat{k} \)
- \( \vec{c} = -\alpha \hat{i} - 2 \hat{j} + \hat{k} \)

### Step 2: Use the first condition \( \vec{a} \cdot \vec{b} = -1 \)
Calculate the dot product:
\[ \vec{a} \cdot \vec{b} = (\hat{i} - \alpha \hat{j} + \beta \hat{k}) \cdot (3 \hat{i} + \beta \hat{j} - \alpha \hat{k}) \]

Calculating the dot product:
\[ = 1 \cdot 3 + (-\alpha) \cdot \beta + \beta \cdot (-\alpha) \]
\[ = 3 - \alpha \beta - \alpha \beta \]
\[ = 3 - 2\alpha \beta \]
Setting this equal to -1:
\[ 3 - 2\alpha \beta = -1 \]
\[ 2\alpha \beta = 4 \quad \Rightarrow \quad \alpha \beta = 2 \quad \text{(Equation 1)} \]

### Step 3: Use the second condition \( \vec{b} \cdot \vec{c} = 10 \)
Calculate the dot product:
\[ \vec{b} \cdot \vec{c} = (3 \hat{i} + \beta \hat{j} - \alpha \hat{k}) \cdot (-\alpha \hat{i} - 2 \hat{j} + \hat{k}) \]

Calculating the dot product:
\[ = 3 \cdot (-\alpha) + \beta \cdot (-2) + (-\alpha) \cdot 1 \]
\[ = -3\alpha - 2\beta - \alpha \]
\[ = -4\alpha - 2\beta \]
Setting this equal to 10:
\[ -4\alpha - 2\beta = 10 \]
Dividing the entire equation by -2:
\[ 2\alpha + \beta = -5 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
We have two equations:
1. \( \alpha \beta = 2 \)
2. \( 2\alpha + \beta = -5 \)

From Equation 2, express \( \beta \) in terms of \( \alpha \):
\[ \beta = -5 - 2\alpha \]

Substituting this into Equation 1:
\[ \alpha(-5 - 2\alpha) = 2 \]
\[ -5\alpha - 2\alpha^2 = 2 \]
Rearranging gives:
\[ 2\alpha^2 + 5\alpha + 2 = 0 \]

### Step 5: Factor the quadratic equation
Factoring:
\[ (2\alpha + 1)(\alpha + 2) = 0 \]
Thus, the solutions for \( \alpha \) are:
\[ \alpha = -\frac{1}{2} \quad \text{or} \quad \alpha = -2 \]

### Step 6: Find corresponding \( \beta \) values
For \( \alpha = -2 \):
\[ \beta = \frac{2}{-2} = -1 \]

For \( \alpha = -\frac{1}{2} \):
\[ \beta = 2 \quad \text{(not an integer, discard)} \]

Thus, we have:
\[ \alpha = -2, \quad \beta = -1 \]

### Step 7: Substitute back to find vectors
Substituting \( \alpha \) and \( \beta \) into the vectors:
\[ \vec{a} = \hat{i} + 2\hat{j} - \hat{k} \]
\[ \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \]
\[ \vec{c} = 2\hat{i} - 2\hat{j} + \hat{k} \]

### Step 8: Calculate \( \vec{a} \times \vec{b} \)
Using the determinant:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}(2 \cdot 2 - (-1) \cdot (-1)) - \hat{j}(1 \cdot 2 - 3 \cdot (-1)) + \hat{k}(1 \cdot (-1) - 2 \cdot 3) \]
\[ = \hat{i}(4 - 1) - \hat{j}(2 + 3) + \hat{k}(-1 - 6) \]
\[ = 3\hat{i} - 5\hat{j} - 7\hat{k} \]

### Step 9: Calculate \( (\vec{a} \times \vec{b}) \cdot \vec{c} \)
Now, calculate the dot product:
\[ (3\hat{i} - 5\hat{j} - 7\hat{k}) \cdot (2\hat{i} - 2\hat{j} + \hat{k}) \]
Calculating:
\[ = 3 \cdot 2 + (-5) \cdot (-2) + (-7) \cdot 1 \]
\[ = 6 + 10 - 7 = 9 \]

### Final Answer
Thus, the value of \( (\vec{a} \times \vec{b}) \cdot \vec{c} \) is:
\[ \boxed{9} \]