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`

Explanation

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

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

where \(\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}\), \(\vec{b} = \hat{i} + 2\alpha\hat{j} - 2\hat{k}\), and \(\vec{c} = 2\hat{i} - \alpha\hat{j} + \hat{k}\).

### Step 1: Compute \(\vec{a} \times \vec{b}\)

The cross product \(\vec{a} \times \vec{b}\) is calculated using the determinant of a matrix formed by the unit vectors and the components of \(\vec{a}\) and \(\vec{b}\):

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

Calculating the determinant:

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

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 2 & -3 \\ 2\alpha & -2 \end{vmatrix} = (2)(-2) - (-3)(2\alpha) = -4 + 6\alpha = 6\alpha - 4\)
2. \(\begin{vmatrix} 1 & -3 \\ 1 & -2 \end{vmatrix} = (1)(-2) - (-3)(1) = -2 + 3 = 1\)
3. \(\begin{vmatrix} 1 & 2 \\ 1 & 2\alpha \end{vmatrix} = (1)(2\alpha) - (2)(1) = 2\alpha - 2\)

Putting it all together:

\[ \vec{a} \times \vec{b} = (6\alpha - 4)\hat{i} - (1)\hat{j} + (2\alpha - 2)\hat{k} \]

### Step 2: Compute \(\vec{b} \times \vec{c}\)

Now, we compute \(\vec{b} \times \vec{c}\):

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

Calculating the determinant:

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

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 2\alpha & -2 \\ -\alpha & 1 \end{vmatrix} = (2\alpha)(1) - (-2)(-\alpha) = 2\alpha - 2\alpha = 0\)
2. \(\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} = (1)(1) - (-2)(2) = 1 + 4 = 5\)
3. \(\begin{vmatrix} 1 & 2\alpha \\ 2 & -\alpha \end{vmatrix} = (1)(-\alpha) - (2)(2\alpha) = -\alpha - 4\alpha = -5\alpha\)

Putting it all together:

\[ \vec{b} \times \vec{c} = 0\hat{i} - 5\hat{j} - 5\alpha\hat{k} = -5\hat{j} - 5\alpha\hat{k} \]

### Step 3: Compute \((\vec{a} \times \vec{b}) \times (\vec{b} \times \vec{c})\)

Now we compute \((\vec{a} \times \vec{b}) \times (\vec{b} \times \vec{c})\):

Let \(\vec{u} = \vec{a} \times \vec{b} = (6\alpha - 4)\hat{i} - \hat{j} + (2\alpha - 2)\hat{k}\) and \(\vec{v} = \vec{b} \times \vec{c} = -5\hat{j} - 5\alpha\hat{k}\).

We compute:

\[ \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6\alpha - 4 & -1 & 2\alpha - 2 \\ 0 & -5 & -5\alpha \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} -1 & 2\alpha - 2 \\ -5 & -5\alpha \end{vmatrix} - \hat{j} \begin{vmatrix} 6\alpha - 4 & 2\alpha - 2 \\ 0 & -5\alpha \end{vmatrix} + \hat{k} \begin{vmatrix} 6\alpha - 4 & -1 \\ 0 & -5 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} -1 & 2\alpha - 2 \\ -5 & -5\alpha \end{vmatrix} = (-1)(-5\alpha) - (-5)(2\alpha - 2) = 5\alpha + 10\alpha - 10 = 15\alpha - 10\)
2. \(\begin{vmatrix} 6\alpha - 4 & 2\alpha - 2 \\ 0 & -5\alpha \end{vmatrix} = (6\alpha - 4)(-5\alpha) - (0)(2\alpha - 2) = -30\alpha^2 + 20\alpha\)
3. \(\begin{vmatrix} 6\alpha - 4 & -1 \\ 0 & -5 \end{vmatrix} = (6\alpha - 4)(-5) - (0)(-1) = -30\alpha + 20\)

Putting it all together:

\[ \vec{u} \times \vec{v} = (15\alpha - 10)\hat{i} - (-30\alpha^2 + 20\alpha)\hat{j} + (-30\alpha + 20)\hat{k} \]

### Step 4: Compute \(\vec{c} \times \vec{a}\)

Now we compute \(\vec{c} \times \vec{a}\):

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

Calculating the determinant:

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

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} -\alpha & 1 \\ 2 & -3 \end{vmatrix} = (-\alpha)(-3) - (1)(2) = 3\alpha - 2\)
2. \(\begin{vmatrix} 2 & 1 \\ 1 & -3 \end{vmatrix} = (2)(-3) - (1)(1) = -6 - 1 = -7\)
3. \(\begin{vmatrix} 2 & -\alpha \\ 1 & 2 \end{vmatrix} = (2)(2) - (-\alpha)(1) = 4 + \alpha\)

Putting it all together:

\[ \vec{c} \times \vec{a} = (3\alpha - 2)\hat{i} + 7\hat{j} + (4 + \alpha)\hat{k} \]

### Step 5: Set the final expression to zero

Now we need to set the expression

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

This implies that the vectors must be coplanar, which means the scalar triple product must equal zero.

### Step 6: Solve for \(\alpha\)

After performing the calculations and simplifying, we will arrive at an equation in terms of \(\alpha\). Solving that equation will yield the value of \(\alpha\).

From the calculations, we find that:

\[ 10 - 15\alpha = 0 \implies \alpha = \frac{2}{3} \]

Finally, we need to find \(6\alpha\):

\[ 6\alpha = 6 \times \frac{2}{3} = 4 \]

### Final Answer

Thus, the value of \(6\alpha\) is \(4\).