Find the shortest distance and the vector equation of the line of shortest distance between the lines given by `r=(3hat(i)+8hat(j)+3hat(k))+lambda(3hat(i)-hat(j)+hat(k)) and r=(-3hat(i)-7hat(j)+6hat(k))+mu(-3hat(i)+2hat(j)+4hat(k))`.

Answer

To find the shortest distance and the vector equation of the line of shortest distance between the given lines, we will follow these steps:

### Step 1: Identify the lines and their components
The lines are given in vector form:
1. Line 1: \( \mathbf{r_1} = (3\hat{i} + 8\hat{j} + 3\hat{k}) + \lambda(3\hat{i} - \hat{j} + \hat{k}) \)
- Position vector \( \mathbf{a_1} = 3\hat{i} + 8\hat{j} + 3\hat{k} \)
- Direction vector \( \mathbf{b_1} = 3\hat{i} - \hat{j} + \hat{k} \)

2. Line 2: \( \mathbf{r_2} = (-3\hat{i} - 7\hat{j} + 6\hat{k}) + \mu(-3\hat{i} + 2\hat{j} + 4\hat{k}) \)
- Position vector \( \mathbf{a_2} = -3\hat{i} - 7\hat{j} + 6\hat{k} \)
- Direction vector \( \mathbf{b_2} = -3\hat{i} + 2\hat{j} + 4\hat{k} \)

### Step 2: Check if the lines are parallel or skew
To check if the lines are parallel, we need to see if the direction vectors \( \mathbf{b_1} \) and \( \mathbf{b_2} \) are proportional. We can do this by checking if the ratios of their components are equal:

- For \( \mathbf{b_1} = (3, -1, 1) \) and \( \mathbf{b_2} = (-3, 2, 4) \):
- The ratios \( \frac{3}{-3}, \frac{-1}{2}, \frac{1}{4} \) are not equal, hence the lines are not parallel and must be skew lines.

### Step 3: Use the formula for the shortest distance between skew lines
The formula for the shortest distance \( d \) between two skew lines is given by:

\[ d = \frac{|\mathbf{b_1} \times \mathbf{b_2} \cdot (\mathbf{a_2} - \mathbf{a_1})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]

### Step 4: Calculate \( \mathbf{b_1} \times \mathbf{b_2} \)
We calculate the cross product \( \mathbf{b_1} \times \mathbf{b_2} \):

\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i}((-1)(4) - (1)(2)) - \hat{j}((3)(4) - (1)(-3)) + \hat{k}((3)(2) - (-1)(-3)) \]
\[ = \hat{i}(-4 - 2) - \hat{j}(12 + 3) + \hat{k}(6 - 3) \]
\[ = -6\hat{i} - 15\hat{j} + 3\hat{k} \]

### Step 5: Calculate \( \mathbf{a_2} - \mathbf{a_1} \)
Now we find \( \mathbf{a_2} - \mathbf{a_1} \):

\[ \mathbf{a_2} - \mathbf{a_1} = (-3\hat{i} - 7\hat{j} + 6\hat{k}) - (3\hat{i} + 8\hat{j} + 3\hat{k}) \]
\[ = (-3 - 3)\hat{i} + (-7 - 8)\hat{j} + (6 - 3)\hat{k} \]
\[ = -6\hat{i} - 15\hat{j} + 3\hat{k} \]

### Step 6: Calculate the dot product \( \mathbf{b_1} \times \mathbf{b_2} \cdot (\mathbf{a_2} - \mathbf{a_1}) \)
Now we compute the dot product:

\[ \mathbf{b_1} \times \mathbf{b_2} \cdot (\mathbf{a_2} - \mathbf{a_1}) = (-6\hat{i} - 15\hat{j} + 3\hat{k}) \cdot (-6\hat{i} - 15\hat{j} + 3\hat{k}) \]
\[ = (-6)(-6) + (-15)(-15) + (3)(3) \]
\[ = 36 + 225 + 9 = 270 \]

### Step 7: Calculate the magnitude \( |\mathbf{b_1} \times \mathbf{b_2}| \)
Now we calculate the magnitude of \( \mathbf{b_1} \times \mathbf{b_2} \):

\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{(-6)^2 + (-15)^2 + (3)^2} \]
\[ = \sqrt{36 + 225 + 9} = \sqrt{270} \]

### Step 8: Calculate the shortest distance \( d \)
Now we can substitute these values into the distance formula:

\[ d = \frac{270}{\sqrt{270}} = \sqrt{270} \]

### Step 9: Simplify \( \sqrt{270} \)
We can simplify \( \sqrt{270} \):

\[ \sqrt{270} = \sqrt{9 \times 30} = 3\sqrt{30} \]

### Step 10: Find the vector equation of the line of shortest distance
The vector equation of the line of shortest distance can be found using the point on each line and the direction vector of the shortest line which is \( \mathbf{b_1} \times \mathbf{b_2} \).

Let \( \mathbf{p_1} = (3, 8, 3) \) and \( \mathbf{p_2} = (-3, -7, 6) \). The direction vector of the line of shortest distance is \( \mathbf{b_1} \times \mathbf{b_2} \).

The vector equation of the line can be represented as:

\[ \mathbf{r} = \mathbf{p_1} + t(\mathbf{b_1} \times \mathbf{b_2}) \quad \text{and} \quad \mathbf{r} = \mathbf{p_2} + s(\mathbf{b_1} \times \mathbf{b_2}) \]

### Final Answer
The shortest distance between the lines is \( 3\sqrt{30} \) and the vector equation of the line of shortest distance can be expressed as:

\[ \mathbf{r} = (3\hat{i} + 8\hat{j} + 3\hat{k}) + t(-6\hat{i} - 15\hat{j} + 3\hat{k}) \]