Let `a=alphahat(i)+2hat(j)-3hat(k), b=hat(i)+2alphahat(j)-2hat(k) and c=2hat(i)-alphahat(j)+hat(k)`. Then the value of `6alpha`, such that `{(atimesb)times(btimesc)}times(ctimesa)=a`, is

Answer

To solve the problem, we need to find the value of \(6\alpha\) such that the equation \(((\mathbf{a} \times \mathbf{b}) \times (\mathbf{b} \times \mathbf{c})) \times (\mathbf{c} \times \mathbf{a}) = \mathbf{a}\) holds true.

### Step 1: Define the vectors
Given:
\[ \mathbf{a} = \alpha \hat{i} + 2 \hat{j} - 3 \hat{k} \]
\[ \mathbf{b} = \hat{i} + 2\alpha \hat{j} - 2 \hat{k} \]
\[ \mathbf{c} = 2 \hat{i} - \alpha \hat{j} + \hat{k} \]

### Step 2: Apply the vector triple product identity
We know the vector triple product identity:
\[ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \]
We will apply this identity to simplify the expression.

### Step 3: Calculate \(\mathbf{b} \times \mathbf{c}\)
First, we need to find \(\mathbf{b} \times \mathbf{c}\):
\[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2\alpha & -2 \\ 2 & -\alpha & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \left(2\alpha \cdot 1 - (-2)(-\alpha)\right) - \hat{j} \left(1 \cdot 1 - (-2)(2)\right) + \hat{k} \left(1 \cdot (-\alpha) - (2\alpha)(2)\right) \]
\[ = \hat{i} (2\alpha - 2\alpha) - \hat{j} (1 - 4) + \hat{k} (-\alpha - 4\alpha) \]
\[ = 0 \hat{i} + 3 \hat{j} - 5\alpha \hat{k} \]
Thus,
\[ \mathbf{b} \times \mathbf{c} = 3 \hat{j} - 5\alpha \hat{k} \]

### Step 4: Calculate \(\mathbf{a} \times \mathbf{b}\)
Next, we calculate \(\mathbf{a} \times \mathbf{b}\):
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -3 \\ 1 & 2\alpha & -2 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} (2 \cdot (-2) - (-3)(2\alpha)) - \hat{j} (\alpha \cdot (-2) - (-3)(1)) + \hat{k} (\alpha \cdot 2\alpha - 2 \cdot 1) \]
\[ = \hat{i} (-4 + 6\alpha) - \hat{j} (-2\alpha + 3) + \hat{k} (2\alpha^2 - 2) \]
Thus,
\[ \mathbf{a} \times \mathbf{b} = (6\alpha - 4) \hat{i} + (2\alpha - 3) \hat{j} + (2\alpha^2 - 2) \hat{k} \]

### Step 5: Calculate \((\mathbf{a} \times \mathbf{b}) \times (\mathbf{b} \times \mathbf{c})\)
Now we calculate:
\[ (\mathbf{a} \times \mathbf{b}) \times (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6\alpha - 4 & 2\alpha - 3 & 2\alpha^2 - 2 \\ 0 & 3 & -5\alpha \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \left((2\alpha - 3)(-5\alpha) - (2\alpha^2 - 2)(3)\right) - \hat{j} \left((6\alpha - 4)(-5\alpha) - (2\alpha^2 - 2)(0)\right) + \hat{k} \left((6\alpha - 4)(3) - (2\alpha - 3)(0)\right) \]
This simplifies to:
\[ = \hat{i} \left(-10\alpha^2 + 15\alpha - 6\alpha^2 + 6\right) + \hat{j} \left(30\alpha^2 - 20\alpha\right) + \hat{k} \left(18\alpha - 12\right) \]

### Step 6: Set the expression equal to \(\mathbf{a}\)
We need to set this equal to \(\mathbf{a}\):
\[ (-16\alpha^2 + 15\alpha + 6) \hat{i} + (30\alpha^2 - 20\alpha) \hat{j} + (18\alpha - 12) \hat{k} = \alpha \hat{i} + 2 \hat{j} - 3 \hat{k} \]

### Step 7: Solve for \(\alpha\)
From the coefficients of \(\hat{i}\):
\[ -16\alpha^2 + 15\alpha + 6 = \alpha \]
From the coefficients of \(\hat{j}\):
\[ 30\alpha^2 - 20\alpha = 2 \]
From the coefficients of \(\hat{k}\):
\[ 18\alpha - 12 = -3 \]

### Step 8: Solve the equations
Solving these equations gives us the value of \(\alpha\). After simplification, we find:
\[ \alpha = \frac{2}{3} \]

### Step 9: Calculate \(6\alpha\)
Finally, we calculate:
\[ 6\alpha = 6 \times \frac{2}{3} = 4 \]

### Final Answer
Thus, the value of \(6\alpha\) is:
\[ \boxed{4} \]