Find the shortest distance between the lines given by
`(x-2)/(3)=(y-5)/(-2)=(z-1)/(1)and(x+1)/(2)=(y+2)/(-6)=(z-3)/(1)`

Options

A. `sqrt29` units
B. `(33)/(sqrt213)` units
C. `2sqrt2` units
D. `20sqrt33` units

Answer

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

1. \((x-2)/(3) = (y-5)/(-2) = (z-1)/(1)\)
2. \((x+1)/(2) = (y+2)/(-6) = (z-3)/(1)\)

we can express these lines in vector form.

### Step 1: Write the lines in vector form

For the first line:
- Point on the line \(A_1 = (2, 5, 1)\)
- Direction vector \(B_1 = (3, -2, 1)\)

The vector form can be written as:
\[ \mathbf{r_1} = \mathbf{A_1} + \lambda \mathbf{B_1} = (2, 5, 1) + \lambda (3, -2, 1) \]

For the second line:
- Point on the line \(A_2 = (-1, -2, 3)\)
- Direction vector \(B_2 = (2, -6, 1)\)

The vector form can be written as:
\[ \mathbf{r_2} = \mathbf{A_2} + \mu \mathbf{B_2} = (-1, -2, 3) + \mu (2, -6, 1) \]

### Step 2: Calculate the cross product of the direction vectors

To find the shortest distance between the two lines, we need to calculate the cross product \(B_1 \times B_2\).

\[ B_1 = (3, -2, 1), \quad B_2 = (2, -6, 1) \]

Calculating the cross product:
\[ B_1 \times B_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 1 \\ 2 & -6 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i} \left((-2)(1) - (1)(-6)\right) - \mathbf{j} \left((3)(1) - (1)(2)\right) + \mathbf{k} \left((3)(-6) - (-2)(2)\right) \]
\[ = \mathbf{i} (-2 + 6) - \mathbf{j} (3 - 2) + \mathbf{k} (-18 + 4) \]
\[ = 4\mathbf{i} - 1\mathbf{j} - 14\mathbf{k} \]
Thus,
\[ B_1 \times B_2 = (4, -1, -14) \]

### Step 3: Calculate the vector \(A_2 - A_1\)

Now, we need to find the vector from point \(A_1\) to point \(A_2\):
\[ A_2 - A_1 = (-1 - 2, -2 - 5, 3 - 1) = (-3, -7, 2) \]

### Step 4: Calculate the shortest distance using the formula

The formula for the shortest distance \(d\) between two skew lines is given by:
\[ d = \frac{|(A_2 - A_1) \cdot (B_1 \times B_2)|}{|B_1 \times B_2|} \]

Calculating the dot product:
\[ (A_2 - A_1) \cdot (B_1 \times B_2) = (-3, -7, 2) \cdot (4, -1, -14) \]
\[ = (-3)(4) + (-7)(-1) + (2)(-14) = -12 + 7 - 28 = -33 \]
Taking the absolute value:
\[ |A_2 - A_1 \cdot (B_1 \times B_2)| = 33 \]

Next, we need to calculate the magnitude of \(B_1 \times B_2\):
\[ |B_1 \times B_2| = \sqrt{4^2 + (-1)^2 + (-14)^2} = \sqrt{16 + 1 + 196} = \sqrt{213} \]

### Step 5: Final calculation of the shortest distance

Now substituting back into the formula:
\[ d = \frac{33}{\sqrt{213}} \]

### Conclusion

The shortest distance between the two lines is:
\[ \frac{33}{\sqrt{213}} \]