Find the distance between the lines `L_(1) and L_(2)` given by :
`vec(r) = hat(i) + 2 hat(j) - 4 hat(k) + lambda (2 hat(i) + 3 hat(j) + 6 hat(k))`
and `vec(r) = 2 hat(i) + 3 hat(j) - 5 hat(k) + mu (2 hat(i) + 3 hat(j) + 6 hat(k))`.

Answer

To find the distance between the two given lines \( L_1 \) and \( L_2 \), we can follow these steps:

### Step 1: Identify the lines and their components
The lines are given in vector form as:
- \( L_1: \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \)
- \( L_2: \vec{r} = 2\hat{i} + 3\hat{j} - 5\hat{k} + \mu(2\hat{i} + 3\hat{j} + 6\hat{k}) \)

From this, we can identify:
- Point \( A \) on \( L_1 \): \( \vec{a} = \hat{i} + 2\hat{j} - 4\hat{k} \)
- Point \( B \) on \( L_2 \): \( \vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k} \)
- Direction vector for both lines: \( \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \)

### Step 2: Check if the lines are parallel
Since both lines have the same direction vector \( \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \), we conclude that \( L_1 \) and \( L_2 \) are parallel.

### Step 3: Calculate the vector connecting points A and B
To find the distance between the two parallel lines, we need to calculate the vector \( \vec{c} - \vec{a} \):
\[ \vec{c} - \vec{a} = \vec{b} - \vec{a} = (2 - 1)\hat{i} + (3 - 2)\hat{j} + (-5 + 4)\hat{k} = \hat{i} + \hat{j} - \hat{k} \]

### Step 4: Use the formula for the distance between parallel lines
The distance \( d \) between two parallel lines can be calculated using the formula:
\[ d = \frac{|(\vec{c} - \vec{a}) \cdot (\vec{b})|}{|\vec{b}|} \]
Where \( \vec{b} \) is the direction vector.

### Step 5: Calculate the cross product
We need to calculate the cross product \( (\vec{c} - \vec{a}) \times \vec{b} \):
\[ \vec{c} - \vec{a} = \hat{i} + \hat{j} - \hat{k} \]
\[ \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \]

Using the determinant to find the cross product:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i}(1 \cdot 6 - (-1) \cdot 3) - \hat{j}(1 \cdot 6 - (-1) \cdot 2) + \hat{k}(1 \cdot 3 - 1 \cdot 2) \]
\[ = \hat{i}(6 + 3) - \hat{j}(6 + 2) + \hat{k}(3 - 2) \]
\[ = 9\hat{i} - 8\hat{j} + 1\hat{k} \]

### Step 6: Calculate the magnitude of the cross product
Now we find the magnitude of the cross product:
\[ |\vec{c} - \vec{a} \times \vec{b}| = \sqrt{9^2 + (-8)^2 + 1^2} = \sqrt{81 + 64 + 1} = \sqrt{146} \]

### Step 7: Calculate the magnitude of the direction vector
Next, we calculate the magnitude of the direction vector \( \vec{b} \):
\[ |\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

### Step 8: Calculate the distance
Finally, we can find the distance \( d \):
\[ d = \frac{|\vec{c} - \vec{a} \times \vec{b}|}{|\vec{b}|} = \frac{\sqrt{146}}{7} \]

### Conclusion
The distance between the lines \( L_1 \) and \( L_2 \) is:
\[ d = \frac{\sqrt{146}}{7} \]