Find the shortest distance between the given line
`vec(r )=(hat(i) +2hat(j) +hat(k)) + lambda(hat(i)-hat(j) + hat(k))`
`vec(r )=(2hat(i) -hat(j) -hat(k)) + lambda(2hat(i) + hat(j) +2hat(k))`

Answer

To find the shortest distance between the two given lines, we can follow these steps:

### Step 1: Identify the lines
The first line is given by:
\[ \vec{r_1} = \hat{i} + 2\hat{j} + \hat{k} + \lambda(\hat{i} - \hat{j} + \hat{k}) \]
The second line is given by:
\[ \vec{r_2} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k}) \]

### Step 2: Extract points and direction vectors
From the first line, we can identify:
- Point \(A_1 = \hat{i} + 2\hat{j} + \hat{k}\)
- Direction vector \(B_1 = \hat{i} - \hat{j} + \hat{k}\)

From the second line, we can identify:
- Point \(A_2 = 2\hat{i} - \hat{j} - \hat{k}\)
- Direction vector \(B_2 = 2\hat{i} + \hat{j} + 2\hat{k}\)

### Step 3: Find the vector between the two points
Calculate the vector \(A_2 - A_1\):
\[ A_2 - A_1 = (2\hat{i} - \hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) = (2 - 1)\hat{i} + (-1 - 2)\hat{j} + (-1 - 1)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k} \]

### Step 4: Calculate the cross product of the direction vectors
Now we need to find the cross product \(B_1 \times B_2\):
\[ B_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \quad B_2 = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} \]
Using the determinant method:
\[ B_1 \times B_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}((-1)(2) - (1)(1)) - \hat{j}((1)(2) - (1)(2)) + \hat{k}((1)(1) - (-1)(2)) \]
\[ = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 + 2) = -3\hat{i} + 0\hat{j} + 3\hat{k} = -3\hat{i} + 3\hat{k} \]

### Step 5: Find the magnitude of the cross product
\[ |B_1 \times B_2| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{9 + 0 + 9} = \sqrt{18} = 3\sqrt{2} \]

### Step 6: Calculate the shortest distance
The formula for the shortest distance \(d\) between two skew lines is given by:
\[ d = \frac{|(A_2 - A_1) \cdot (B_1 \times B_2)|}{|B_1 \times B_2|} \]
Calculating the dot product:
\[ A_2 - A_1 = \hat{i} - 3\hat{j} - 2\hat{k} \]
\[ A_2 - A_1 \cdot (B_1 \times B_2) = (\hat{i} - 3\hat{j} - 2\hat{k}) \cdot (-3\hat{i} + 3\hat{k}) = (-3)(1) + (0)(-3) + (3)(-2) = -3 + 0 - 6 = -9 \]

Now substituting 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:
\[ \frac{3\sqrt{2}}{2} \]