Find the shortest distance between the given line
`vec(r ) =(6hat(i) +3hat(k) ) + lambda(2hat(i) -hat(j) +4hat(k))`
`vec(r )=(-9hat(i) +hat(j) -10hat(k)) + mu (4hat(i) +hat(j) +6hat(k))`

Answer

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

### Step 1: Identify the lines
The two lines are given in vector form:
1. \( \vec{r_1} = (6 \hat{i} + 3 \hat{k}) + \lambda(2 \hat{i} - \hat{j} + 4 \hat{k}) \)
2. \( \vec{r_2} = (-9 \hat{i} + \hat{j} - 10 \hat{k}) + \mu(4 \hat{i} + \hat{j} + 6 \hat{k}) \)

### Step 2: Extract points and direction vectors
From the equations:
- For line 1:
- Point \( \vec{a_1} = 6 \hat{i} + 3 \hat{k} \)
- Direction vector \( \vec{b_1} = 2 \hat{i} - \hat{j} + 4 \hat{k} \)

- For line 2:
- Point \( \vec{a_2} = -9 \hat{i} + \hat{j} - 10 \hat{k} \)
- Direction vector \( \vec{b_2} = 4 \hat{i} + \hat{j} + 6 \hat{k} \)

### Step 3: Calculate \( \vec{a_2} - \vec{a_1} \)
\[ \vec{a_2} - \vec{a_1} = (-9 \hat{i} + \hat{j} - 10 \hat{k}) - (6 \hat{i} + 3 \hat{k}) \]
\[ = (-9 - 6) \hat{i} + (1 - 0) \hat{j} + (-10 - 3) \hat{k} \]
\[ = -15 \hat{i} + 1 \hat{j} - 13 \hat{k} \]

### Step 4: Compute \( \vec{b_1} \times \vec{b_2} \)
Using the determinant to find the cross product:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 4 \\ 4 & 1 & 6 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 4 \\ 1 & 6 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 4 & 6 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix} \]
Calculating each determinant:
\[ = \hat{i}((-1)(6) - (1)(4)) - \hat{j}((2)(6) - (4)(4)) + \hat{k}((2)(1) - (-1)(4)) \]
\[ = \hat{i}(-6 - 4) - \hat{j}(12 - 16) + \hat{k}(2 + 4) \]
\[ = -10 \hat{i} + 4 \hat{j} + 6 \hat{k} \]

### Step 5: Find the magnitude of \( \vec{b_1} \times \vec{b_2} \)
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(-10)^2 + 4^2 + 6^2} = \sqrt{100 + 16 + 36} = \sqrt{152} \]

### Step 6: Use the formula for shortest distance
The formula for the shortest distance \( d \) between two skew lines is given by:
\[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]

### Step 7: Calculate the dot product
Calculating \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \):
\[ (-15 \hat{i} + 1 \hat{j} - 13 \hat{k}) \cdot (-10 \hat{i} + 4 \hat{j} + 6 \hat{k}) \]
\[ = (-15)(-10) + (1)(4) + (-13)(6) \]
\[ = 150 + 4 - 78 = 76 \]

### Step 8: Substitute values into the distance formula
\[ d = \frac{|76|}{\sqrt{152}} = \frac{76}{\sqrt{152}} = \frac{76}{\sqrt{4 \cdot 38}} = \frac{76}{2\sqrt{38}} = \frac{38}{\sqrt{38}} = \sqrt{38} \]

### Final Answer
The shortest distance between the two lines is:
\[ \sqrt{38} \]