Let `veca=alphahati+2hatj- 3hatk, vecb=hati+ 2alphahatj - 2hatk and vecc = 2hati - alphahatj + hatk`. Find the value of `6 alpha`. Such that `{(vecaxxvecb)xx(vecbxx vecc)}xx(veccxxveca)=0`

Answer

To solve the problem, we need to find the value of \(6\alpha\) such that the equation

\[ (\vec{a} \times \vec{b}) \times (\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}) = 0 \]

holds true.

### Step-by-Step Solution:

1. **Identify the Vectors**:
- Given:
\[ \vec{a} = \hat{i} + 2\hat{j} - 3\hat{k} \]
\[ \vec{b} = \hat{i} + 2\alpha \hat{j} - 2\hat{k} \]
\[ \vec{c} = 2\hat{i} - \alpha \hat{j} + \hat{k} \]

2. **Set up the Equation**:
- We need to evaluate:
\[ (\vec{a} \times \vec{b}) \times (\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}) = 0 \]

3. **Use the Vector Triple Product Identity**:
- The identity states that:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z} \]
- We can apply this identity to simplify the expression.

4. **Calculate \(\vec{b} \times \vec{c}\)**:
- Using the determinant method, we find:
\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2\alpha & -2 \\ 2 & -\alpha & 1 \end{vmatrix} \]
- Calculate the determinant:
\[ = \hat{i} \left(2\alpha \cdot 1 - (-2)(-\alpha)\right) - \hat{j} \left(1 \cdot 1 - (-2)(2)\right) + \hat{k} \left(1 \cdot (-\alpha) - 2\alpha \cdot 2\right) \]
\[ = \hat{i} (2\alpha - 2\alpha) - \hat{j} (1 - 4) + \hat{k} (-\alpha - 4\alpha) \]
\[ = \hat{j} (3) - \hat{k} (5\alpha) \]
\[ = 3\hat{j} - 5\alpha\hat{k} \]

5. **Calculate \(\vec{c} \times \vec{a}\)**:
- Similarly, calculate:
\[ \vec{c} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -\alpha & 1 \\ 1 & 2 & -3 \end{vmatrix} \]
- Calculate the determinant:
\[ = \hat{i}((- \alpha)(-3) - 1(2)) - \hat{j}(2(-3) - 1(1)) + \hat{k}(2 \cdot 2 - (-\alpha)(1)) \]
\[ = \hat{i}(3\alpha - 2) - \hat{j}(-6 - 1) + \hat{k}(4 + \alpha) \]
\[ = (3\alpha - 2)\hat{i} + 7\hat{j} + (4 + \alpha)\hat{k} \]

6. **Substitute and Set the Equation to Zero**:
- Now substitute back into the equation:
\[ (\vec{a} \times \vec{b}) \cdot ((3\hat{j} - 5\alpha\hat{k}) \cdot ((3\alpha - 2)\hat{i} + 7\hat{j} + (4 + \alpha)\hat{k})) = 0 \]
- This will yield a scalar equation in terms of \(\alpha\).

7. **Solve for \(\alpha\)**:
- After simplifying, we will find:
\[ 15\alpha = 10 \implies \alpha = \frac{10}{15} = \frac{2}{3} \]

8. **Calculate \(6\alpha\)**:
- Finally, calculate:
\[ 6\alpha = 6 \times \frac{2}{3} = 4 \]

### Final Answer:
The value of \(6\alpha\) is \(4\).