`vec(r )=(-4hat(i)+4hat(j) +hat(k)) + lambda (hat(i) +hat(j) -hat(k))`
`vec(r)=(-3hat(i) -8hat(j) -3hat(k)) + mu (2hat(i) +3hat(j) +3hat(k))`

Answer

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

1. **Line 1**: \(\vec{r_1} = (-4\hat{i} + 4\hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} - \hat{k})\)
2. **Line 2**: \(\vec{r_2} = (-3\hat{i} - 8\hat{j} - 3\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 3\hat{k})\)

We will use the formula for the shortest distance \(d\) between two skew lines:

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

Where:
- \(\vec{a_1} = -4\hat{i} + 4\hat{j} + \hat{k}\)
- \(\vec{a_2} = -3\hat{i} - 8\hat{j} - 3\hat{k}\)
- \(\vec{b_1} = \hat{i} + \hat{j} - \hat{k}\)
- \(\vec{b_2} = 2\hat{i} + 3\hat{j} + 3\hat{k}\)

### Step 1: Calculate \(\vec{a_2} - \vec{a_1}\)

\[ \vec{a_2} - \vec{a_1} = (-3\hat{i} - 8\hat{j} - 3\hat{k}) - (-4\hat{i} + 4\hat{j} + \hat{k}) \]

Calculating this gives:

\[ \vec{a_2} - \vec{a_1} = (-3 + 4)\hat{i} + (-8 - 4)\hat{j} + (-3 - 1)\hat{k} = \hat{i} - 12\hat{j} - 4\hat{k} \]

### Step 2: Calculate \(\vec{b_1} \times \vec{b_2}\)

Using the determinant method for the cross product:

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

Calculating this determinant:

\[ = \hat{i} \begin{vmatrix} 1 & -1 \\ 3 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} 1 & -1 \\ 3 & 3 \end{vmatrix} = (1)(3) - (-1)(3) = 3 + 3 = 6\)
2. \(\begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (-1)(2) = 3 + 2 = 5\)
3. \(\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (1)(2) = 3 - 2 = 1\)

Putting it all together:

\[ \vec{b_1} \times \vec{b_2} = 6\hat{i} - 5\hat{j} + 1\hat{k} \]

### Step 3: Calculate the magnitude of \(\vec{b_1} \times \vec{b_2}\)

\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{6^2 + (-5)^2 + 1^2} = \sqrt{36 + 25 + 1} = \sqrt{62} \]

### Step 4: Calculate the dot product \((\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})\)

\[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (\hat{i} - 12\hat{j} - 4\hat{k}) \cdot (6\hat{i} - 5\hat{j} + 1\hat{k}) \]

Calculating this gives:

\[ = (1)(6) + (-12)(-5) + (-4)(1) = 6 + 60 - 4 = 62 \]

### Step 5: Substitute into the distance formula

\[ d = \frac{|62|}{\sqrt{62}} = \frac{62}{\sqrt{62}} = \sqrt{62} \]

### Final Answer

The shortest distance between the two lines is:

\[ d = \sqrt{62} \]