Find vector `vec(v)` which is coplanar with the vectors `hat(i) + hat(j) - 2hat(k)` and `hat(i) - 2hat(j) + hat(k)` and is orthogonal to the vector `-2hat(j) + hat(k)` . It is given that the projection of `vec(v)` along the vector `hat(i) + hat(j) + hat(k)` is equal to `6 sqrt(3)`.

Answer

To find the vector \(\vec{v}\) that is coplanar with the vectors \(\hat{i} + \hat{j} - 2\hat{k}\) and \(\hat{i} - 2\hat{j} + \hat{k}\), orthogonal to the vector \(-2\hat{j} + \hat{k}\), and has a projection of \(6\sqrt{3}\) along the vector \(\hat{i} + \hat{j} + \hat{k}\), we can follow these steps:

### Step 1: Define the vectors
Let:
\[ \vec{a} = \hat{i} + \hat{j} - 2\hat{k} \]
\[ \vec{b} = \hat{i} - 2\hat{j} + \hat{k} \]
\[ \vec{c} = -2\hat{j} + \hat{k} \]

### Step 2: Find the cross product of \(\vec{a}\) and \(\vec{b}\)
To find a vector that is coplanar with \(\vec{a}\) and \(\vec{b}\), we can use the cross product:
\[ \vec{n} = \vec{a} \times \vec{b} \]

Calculating \(\vec{n}\):
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 1 & -2 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i} \begin{vmatrix} 1 & -2 \\ -2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} \]

Calculating the 2x2 determinants:
\[ \begin{vmatrix} 1 & -2 \\ -2 & 1 \end{vmatrix} = 1 \cdot 1 - (-2) \cdot (-2) = 1 - 4 = -3 \]
\[ \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} = 1 \cdot 1 - (-2) \cdot 1 = 1 + 2 = 3 \]
\[ \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = 1 \cdot (-2) - 1 \cdot 1 = -2 - 1 = -3 \]

Substituting back:
\[ \vec{n} = -3\hat{i} - 3\hat{j} - 3\hat{k} = -3(\hat{i} + \hat{j} + \hat{k}) \]

### Step 3: Express \(\vec{v}\) in terms of \(\vec{n}\)
Since \(\vec{v}\) is coplanar with \(\vec{a}\) and \(\vec{b}\), we can express \(\vec{v}\) as:
\[ \vec{v} = \alpha \vec{n} \]
where \(\alpha\) is a scalar.

### Step 4: Orthogonality condition
For \(\vec{v}\) to be orthogonal to \(\vec{c}\):
\[ \vec{v} \cdot \vec{c} = 0 \]
Substituting \(\vec{v}\):
\[ \alpha(-3(\hat{i} + \hat{j} + \hat{k})) \cdot (-2\hat{j} + \hat{k}) = 0 \]
Calculating the dot product:
\[ -3\alpha \left(0 + (-2) + 1\right) = 0 \]
This simplifies to:
\[ -3\alpha(-1) = 0 \implies \alpha \neq 0 \]

### Step 5: Projection condition
The projection of \(\vec{v}\) along \(\hat{i} + \hat{j} + \hat{k}\) is given as \(6\sqrt{3}\):
\[ \text{Projection} = \frac{\vec{v} \cdot (\hat{i} + \hat{j} + \hat{k})}{|\hat{i} + \hat{j} + \hat{k}|} \]
Calculating the magnitude:
\[ |\hat{i} + \hat{j} + \hat{k}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
Thus:
\[ \vec{v} \cdot (\hat{i} + \hat{j} + \hat{k}) = 6\sqrt{3} \]
Substituting \(\vec{v}\):
\[ \alpha(-3(\hat{i} + \hat{j} + \hat{k})) \cdot (\hat{i} + \hat{j} + \hat{k}) = 6\sqrt{3} \]
Calculating the dot product:
\[ -3\alpha(3) = 6\sqrt{3} \implies -9\alpha = 6\sqrt{3} \implies \alpha = -\frac{2\sqrt{3}}{3} \]

### Step 6: Final vector \(\vec{v}\)
Substituting \(\alpha\) back:
\[ \vec{v} = -\frac{2\sqrt{3}}{3}(-3(\hat{i} + \hat{j} + \hat{k})) = 2\sqrt{3}(\hat{i} + \hat{j} + \hat{k}) = 2\sqrt{3}\hat{i} + 2\sqrt{3}\hat{j} + 2\sqrt{3}\hat{k} \]

### Final Answer
\[ \vec{v} = 2\sqrt{3}(\hat{i} + \hat{j} + \hat{k}) \]