The shortest distance between the lines `(x-1)/2=(y+8)/-7=(z-4)/5` and `(x-1)/2=(y-2)/1=(z-6)/-3` is

Options

A. `2sqrt3`
B. `5sqrt3`
C. `4sqrt3`
D. `3sqrt3`

Answer

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

1. \(\frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}\) (Line \(l_1\))
2. \(\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3}\) (Line \(l_2\))

we can follow these steps:

### Step 1: Identify Points and Direction Vectors

For line \(l_1\):
- A point on the line \(P_1(1, -8, 4)\)
- Direction vector \(b_1 = (2, -7, 5)\)

For line \(l_2\):
- A point on the line \(P_2(1, 2, 6)\)
- Direction vector \(b_2 = (2, 1, -3)\)

### Step 2: Find the Vector Connecting the Points

The vector connecting points \(P_1\) and \(P_2\) is:
\[ \vec{AC} = P_2 - P_1 = (1 - 1, 2 - (-8), 6 - 4) = (0, 10, 2) \]

### Step 3: Compute the Cross Product of Direction Vectors

Now we compute the cross product \(b_1 \times b_2\):
\[ b_1 = (2, -7, 5), \quad b_2 = (2, 1, -3) \]

Using the determinant:
\[ b_1 \times b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i}((-7)(-3) - (5)(1)) - \hat{j}((2)(-3) - (5)(2)) + \hat{k}((2)(1) - (-7)(2)) \]
\[ = \hat{i}(21 - 5) - \hat{j}(-6 - 10) + \hat{k}(2 + 14) \]
\[ = \hat{i}(16) + \hat{j}(16) + \hat{k}(16) = (16, 16, 16) \]

### Step 4: Calculate the Magnitude of the Cross Product

The magnitude of \(b_1 \times b_2\) is:
\[ |b_1 \times b_2| = \sqrt{16^2 + 16^2 + 16^2} = \sqrt{768} = 16\sqrt{3} \]

### Step 5: Calculate the Shortest Distance

The formula for the shortest distance \(d\) between the two skew lines is given by:
\[ d = \frac{|\vec{AC} \cdot (b_1 \times b_2)|}{|b_1 \times b_2|} \]

Calculating the dot product:
\[ \vec{AC} \cdot (b_1 \times b_2) = (0, 10, 2) \cdot (16, 16, 16) = 0 \cdot 16 + 10 \cdot 16 + 2 \cdot 16 = 160 + 32 = 192 \]

Now substituting into the distance formula:
\[ d = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \]

### Final Answer
Thus, the shortest distance between the lines is:
\[ \boxed{4\sqrt{3}} \]