Find the shortest distance between the given line
`vec(r )=(hat(i) +hat(j)) +lambda(2hat(i) -hat(j) +hat(k))`
`vec(r )=(2hat(i) +hat(j) -hat(k)) + mu (3hat(i) -5hat(j) +2hat(k))`

Answer

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

### Step 1: Identify the points and direction vectors of the lines

The given lines are:

1. \(\vec{r_1} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})\)
2. \(\vec{r_2} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(3\hat{i} - 5\hat{j} + 2\hat{k})\)

From these equations, we can identify:
- For line 1:
- Point \(A_1 = \hat{i} + \hat{j}\)
- Direction vector \(B_1 = 2\hat{i} - \hat{j} + \hat{k}\)

- For line 2:
- Point \(A_2 = 2\hat{i} + \hat{j} - \hat{k}\)
- Direction vector \(B_2 = 3\hat{i} - 5\hat{j} + 2\hat{k}\)

### Step 2: Calculate \(A_2 - A_1\)

Next, we find the vector between the two points \(A_2\) and \(A_1\):

\[ A_2 - A_1 = (2\hat{i} + \hat{j} - \hat{k}) - (\hat{i} + \hat{j}) = (2\hat{i} - \hat{i}) + (\hat{j} - \hat{j}) + (-\hat{k}) = \hat{i} - \hat{k} \]

### Step 3: Calculate the cross product \(B_1 \times B_2\)

Now, we need to find the cross product of the direction vectors \(B_1\) and \(B_2\):

\[ B_1 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}, \quad B_2 = \begin{pmatrix} 3 \\ -5 \\ 2 \end{pmatrix} \]

Using the determinant method to find \(B_1 \times B_2\):

\[ B_1 \times B_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i}((-1)(2) - (1)(-5)) - \hat{j}((2)(2) - (1)(3)) + \hat{k}((2)(-5) - (-1)(3)) \]
\[ = \hat{i}(-2 + 5) - \hat{j}(4 - 3) + \hat{k}(-10 + 3) \]
\[ = 3\hat{i} - 1\hat{j} - 7\hat{k} \]

### Step 4: Calculate the magnitude of the cross product

Now, we find the magnitude of \(B_1 \times B_2\):

\[ |B_1 \times B_2| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9 + 1 + 49} = \sqrt{59} \]

### Step 5: Calculate the shortest distance

The shortest distance \(d\) between the two lines is given by the formula:

\[ d = \frac{|(A_2 - A_1) \cdot (B_1 \times B_2)|}{|B_1 \times B_2|} \]

Calculating the dot product \((A_2 - A_1) \cdot (B_1 \times B_2)\):

\[ A_2 - A_1 = \hat{i} - \hat{k} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \]
\[ B_1 \times B_2 = \begin{pmatrix} 3 \\ -1 \\ -7 \end{pmatrix} \]

Now, the dot product:

\[ (A_2 - A_1) \cdot (B_1 \times B_2) = (1)(3) + (0)(-1) + (-1)(-7) = 3 + 0 + 7 = 10 \]

Now substituting into the distance formula:

\[ d = \frac{|10|}{\sqrt{59}} = \frac{10}{\sqrt{59}} \]

### Final Answer

Thus, the shortest distance between the given lines is:

\[ \frac{10}{\sqrt{59}} \text{ units} \]