Find the shortest distance and the vector equation of the line of shortest distance between the lines given by:
`vec(r) = (3 hat(i) + 8 hat(j) + 3 hat(k) ) + lambda (3 hat(i) - hat(j) + hat(k))` and
`vec(r) = (-3 hat(i) - 7 hat(j) + 6 hat(k)) + mu (-3 hat(i) + 2 hat(j) + 4 hat(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 given lines
The two lines are represented as:
1. Line 1: \(\vec{r_1} = (3 \hat{i} + 8 \hat{j} + 3 \hat{k}) + \lambda (3 \hat{i} - \hat{j} + \hat{k})\)
2. Line 2: \(\vec{r_2} = (-3 \hat{i} - 7 \hat{j} + 6 \hat{k}) + \mu (-3 \hat{i} + 2 \hat{j} + 4 \hat{k})\)

### Step 2: Extract points and direction vectors
From the equations, we can identify:
- Point on Line 1, \( \vec{a_1} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} \)
- Direction vector of Line 1, \( \vec{b_1} = 3 \hat{i} - \hat{j} + \hat{k} \)

- Point on Line 2, \( \vec{a_2} = -3 \hat{i} - 7 \hat{j} + 6 \hat{k} \)
- Direction vector of Line 2, \( \vec{b_2} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k} \)

### Step 3: Calculate \( \vec{a_2} - \vec{a_1} \)
\[ \vec{a_2} - \vec{a_1} = (-3 \hat{i} - 7 \hat{j} + 6 \hat{k}) - (3 \hat{i} + 8 \hat{j} + 3 \hat{k}) = -6 \hat{i} - 15 \hat{j} + 3 \hat{k} \]

### Step 4: Calculate the cross product \( \vec{b_1} \times \vec{b_2} \)
Using the determinant method:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} \]

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

### Step 6: Calculate the shortest distance \( d \)
Using the formula for the shortest distance between two skew lines:
\[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]
Calculating the dot product:
\[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-6 \hat{i} - 15 \hat{j} + 3 \hat{k}) \cdot (-6 \hat{i} - 15 \hat{j} + 3 \hat{k}) \]
\[ = (-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270 \]
Thus,
\[ d = \frac{|270|}{\sqrt{270}} = \sqrt{270} \]

### Step 7: Find the vector equation of the line of shortest distance
The line of shortest distance passes through point \( \vec{a_1} \) with direction \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{r} = \vec{a_1} + \gamma (\vec{b_1} \times \vec{b_2}) \]
Substituting the values:
\[ \vec{r} = (3 \hat{i} + 8 \hat{j} + 3 \hat{k}) + \gamma (-6 \hat{i} - 15 \hat{j} + 3 \hat{k}) \]

### Final Answer
The shortest distance is \( \sqrt{270} \) and the vector equation of the line of shortest distance is:
\[ \vec{r} = (3 - 6\gamma) \hat{i} + (8 - 15\gamma) \hat{j} + (3 + 3\gamma) \hat{k} \]