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

Options

A. `5/sqrt(14)`
B. `3/sqrt(17)`
C. `6/sqrt(14)`
D. `2/sqrt(15)`

Answer

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

\[ \frac{x-1}{2} = \frac{y-1}{3} = \frac{z+2}{6} \]

and

\[ \frac{x-2}{4} = \frac{y+1}{1} = \frac{z-3}{7}, \]

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

\[ d = \frac{| \mathbf{a_2} - \mathbf{a_1} \cdot (\mathbf{n_1} \times \mathbf{n_2}) |}{|\mathbf{n_1} \times \mathbf{n_2}|} \]

where \(\mathbf{a_1}\) and \(\mathbf{a_2}\) are points on the first and second lines, respectively, and \(\mathbf{n_1}\) and \(\mathbf{n_2}\) are the direction ratios of the lines.

### Step 1: Identify points and direction ratios

From the first line, we can extract:
- A point \(\mathbf{a_1} = (1, 1, -2)\)
- Direction ratios \(\mathbf{n_1} = (2, 3, 6)\)

From the second line, we can extract:
- A point \(\mathbf{a_2} = (2, -1, 3)\)
- Direction ratios \(\mathbf{n_2} = (4, 1, 7)\)

### Step 2: Calculate \(\mathbf{a_2} - \mathbf{a_1}\)

Now, we calculate the vector \(\mathbf{a_2} - \mathbf{a_1}\):

\[ \mathbf{a_2} - \mathbf{a_1} = (2 - 1, -1 - 1, 3 - (-2)) = (1, -2, 5) \]

### Step 3: Calculate \(\mathbf{n_1} \times \mathbf{n_2}\)

Next, we compute the cross product \(\mathbf{n_1} \times \mathbf{n_2}\):

\[ \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 6 \\ 4 & 1 & 7 \end{vmatrix} \]

Calculating the determinant, we get:

\[ \mathbf{n_1} \times \mathbf{n_2} = \mathbf{i}(3 \cdot 7 - 6 \cdot 1) - \mathbf{j}(2 \cdot 7 - 6 \cdot 4) + \mathbf{k}(2 \cdot 1 - 3 \cdot 4) \]

\[ = \mathbf{i}(21 - 6) - \mathbf{j}(14 - 24) + \mathbf{k}(2 - 12) \]

\[ = 15\mathbf{i} + 10\mathbf{j} - 10\mathbf{k} = (15, 10, -10) \]

### Step 4: Calculate the dot product

Now we calculate the dot product \((\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{n_1} \times \mathbf{n_2})\):

\[ (1, -2, 5) \cdot (15, 10, -10) = 1 \cdot 15 + (-2) \cdot 10 + 5 \cdot (-10) \]

\[ = 15 - 20 - 50 = -55 \]

Taking the absolute value, we have:

\[ |(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{n_1} \times \mathbf{n_2})| = 55 \]

### Step 5: Calculate \(|\mathbf{n_1} \times \mathbf{n_2}|\)

Next, we calculate the magnitude of \(\mathbf{n_1} \times \mathbf{n_2}\):

\[ |\mathbf{n_1} \times \mathbf{n_2}| = \sqrt{15^2 + 10^2 + (-10)^2} = \sqrt{225 + 100 + 100} = \sqrt{425} = 5\sqrt{17} \]

### Step 6: Calculate the shortest distance \(d\)

Finally, we substitute the values into the distance formula:

\[ d = \frac{55}{5\sqrt{17}} = \frac{11}{\sqrt{17}} \]

### Conclusion

Thus, the shortest distance between the two lines is:

\[ d = \frac{11}{\sqrt{17}}. \]