Find the shortest distance betwee the lines :
`vec(r) = (hati + 2 hatj + hatk ) + lambda ( hati - hatj + hatk)` and
`vec(r) = 2 hati - hatj - hakt + mu (2 hati + hatj + 2 hatk)`.

Answer

To find the shortest distance between the two lines given by the vector equations:

1. **Line L1**: \(\vec{r_1} = \hat{i} + 2\hat{j} + \hat{k} + \lambda(\hat{i} - \hat{j} + \hat{k})\)
2. **Line L2**: \(\vec{r_2} = 2\hat{i} - \hat{j} - \hat{k} + \mu(2\hat{i} + \hat{j} + 2\hat{k})\)

We can denote the points and direction vectors of the lines as follows:

- For Line L1:
- Point \(A = \hat{i} + 2\hat{j} + \hat{k}\)
- Direction vector \( \vec{b} = \hat{i} - \hat{j} + \hat{k}\)

- For Line L2:
- Point \(C = 2\hat{i} - \hat{j} - \hat{k}\)
- Direction vector \( \vec{d} = 2\hat{i} + \hat{j} + 2\hat{k}\)

### Step 1: Find the vector \( \vec{AC} \)
The vector from point A to point C is given by:
\[ \vec{AC} = C - A = (2\hat{i} - \hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) = (2 - 1)\hat{i} + (-1 - 2)\hat{j} + (-1 - 1)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k} \]

### Step 2: Calculate the cross product \( \vec{b} \times \vec{d} \)
Now, we need to find the cross product of the direction vectors \( \vec{b} \) and \( \vec{d} \):
\[ \vec{b} = \hat{i} - \hat{j} + \hat{k}, \quad \vec{d} = 2\hat{i} + \hat{j} + 2\hat{k} \]
Using the determinant method to find the cross product:
\[ \vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
\[ = \hat{i}((-1)(2) - (1)(1)) - \hat{j}((1)(2) - (1)(2)) + \hat{k}((1)(1) - (-1)(2)) \]
\[ = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 + 2) \]
\[ = -3\hat{i} + 0\hat{j} + 3\hat{k} = -3\hat{i} + 3\hat{k} \]

### Step 3: Find the magnitude of the cross product
The magnitude of the cross product is:
\[ |\vec{b} \times \vec{d}| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{9 + 0 + 9} = \sqrt{18} = 3\sqrt{2} \]

### Step 4: Find the shortest distance
The shortest distance \(d\) between the two lines can be calculated using the formula:
\[ d = \frac{|\vec{AC} \cdot (\vec{b} \times \vec{d})|}{|\vec{b} \times \vec{d}|} \]
First, we calculate the dot product \( \vec{AC} \cdot (\vec{b} \times \vec{d}) \):
\[ \vec{AC} = \hat{i} - 3\hat{j} - 2\hat{k}, \quad \vec{b} \times \vec{d} = -3\hat{i} + 3\hat{k} \]
Calculating the dot product:
\[ \vec{AC} \cdot (\vec{b} \times \vec{d}) = (1)(-3) + (-3)(0) + (-2)(3) = -3 + 0 - 6 = -9 \]
Now substituting back into the distance formula:
\[ d = \frac{|-9|}{3\sqrt{2}} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \]

### Final Answer
The shortest distance between the two lines is:
\[ \frac{3\sqrt{2}}{2} \]