The shortest distance between the lines `(x-3)/3=(y-8)/(-1)=(z-3)/1` and `(x+3)/(-3)=(y+7)/2=(z-6)/4` is

Options

A. a. `sqrt(30)`
B. b. `2sqrt(30)`
C. c. `5sqrt(30)`
D. d. `3sqrt(30)`

Answer

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

### Step 1: Identify the equations of the lines
The first line is given by:
\[ \frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1} \]
From this, we can identify:
- Point \( P_1(3, 8, 3) \)
- Direction ratios \( \mathbf{a_1} = (3, -1, 1) \)

The second line is given by:
\[ \frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4} \]
From this, we can identify:
- Point \( P_2(-3, -7, 6) \)
- Direction ratios \( \mathbf{a_2} = (-3, 2, 4) \)

### Step 2: Use the formula for the shortest distance between two skew lines
The formula for the shortest distance \( d \) between two skew lines is given by:
\[ d = \frac{|(\mathbf{P_2} - \mathbf{P_1}) \cdot (\mathbf{a_1} \times \mathbf{a_2})|}{|\mathbf{a_1} \times \mathbf{a_2}|} \]
Where:
- \( \mathbf{P_2} - \mathbf{P_1} = (-3 - 3, -7 - 8, 6 - 3) = (-6, -15, 3) \)

### Step 3: Calculate the cross product \( \mathbf{a_1} \times \mathbf{a_2} \)
\[ \mathbf{a_1} = (3, -1, 1), \quad \mathbf{a_2} = (-3, 2, 4) \]
Using the determinant to find the cross product:
\[ \mathbf{a_1} \times \mathbf{a_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} \]
Calculating this determinant:
\[ = \mathbf{i}((-1)(4) - (1)(2)) - \mathbf{j}((3)(4) - (1)(-3)) + \mathbf{k}((3)(2) - (-1)(-3)) \]
\[ = \mathbf{i}(-4 - 2) - \mathbf{j}(12 + 3) + \mathbf{k}(6 - 3) \]
\[ = -6\mathbf{i} - 15\mathbf{j} + 3\mathbf{k} \]
Thus,
\[ \mathbf{a_1} \times \mathbf{a_2} = (-6, -15, 3) \]

### Step 4: Calculate the magnitude of the cross product
\[ |\mathbf{a_1} \times \mathbf{a_2}| = \sqrt{(-6)^2 + (-15)^2 + (3)^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30} \]

### Step 5: Calculate the dot product \( (\mathbf{P_2} - \mathbf{P_1}) \cdot (\mathbf{a_1} \times \mathbf{a_2}) \)
\[ (\mathbf{P_2} - \mathbf{P_1}) \cdot (\mathbf{a_1} \times \mathbf{a_2}) = (-6, -15, 3) \cdot (-6, -15, 3) \]
Calculating this dot product:
\[ = (-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270 \]

### Step 6: Substitute values into the distance formula
Now substituting back into the distance formula:
\[ d = \frac{|270|}{3\sqrt{30}} = \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = 3\sqrt{30} \]

### Conclusion
The shortest distance between the two lines is:
\[ \boxed{3\sqrt{30}} \]