Find the magnitude of the shortest distance between the lines `(x)/(2)=(y)/(-3)=(z)/(1) and (x-2)/(3)=(y-1)/(-5)=(z+2)/(2)`.

Answer

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

1. \(\frac{x}{2} = \frac{y}{-3} = \frac{z}{1}\)
2. \(\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}\)

we can follow these steps:

### Step 1: Write the lines in vector form
For the first line, we can express it in the vector form:
\[ \mathbf{r_1} = \mathbf{a_1} + \lambda \mathbf{b_1} \]
where \(\mathbf{a_1} = (0, 0, 0)\) (a point on the line) and \(\mathbf{b_1} = (2, -3, 1)\) (the direction ratios).

Thus, the vector form of line 1 is:
\[ \mathbf{r_1} = (0, 0, 0) + \lambda (2, -3, 1) = (2\lambda, -3\lambda, \lambda) \]

For the second line:
\[ \mathbf{r_2} = \mathbf{a_2} + \mu \mathbf{b_2} \]
where \(\mathbf{a_2} = (2, 1, -2)\) and \(\mathbf{b_2} = (3, -5, 2)\).

Thus, the vector form of line 2 is:
\[ \mathbf{r_2} = (2, 1, -2) + \mu (3, -5, 2) = (2 + 3\mu, 1 - 5\mu, -2 + 2\mu) \]

### Step 2: Calculate the cross product of the direction vectors
Next, we need to find \(\mathbf{b_1} \times \mathbf{b_2}\):
\[ \mathbf{b_1} = (2, -3, 1), \quad \mathbf{b_2} = (3, -5, 2) \]
Calculating the cross product:
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 1 \\ 3 & -5 & 2 \end{vmatrix} \]
Calculating the determinant, we get:
\[ = \mathbf{i}((-3)(2) - (1)(-5)) - \mathbf{j}((2)(2) - (1)(3)) + \mathbf{k}((2)(-5) - (-3)(3)) \]
\[ = \mathbf{i}(-6 + 5) - \mathbf{j}(4 - 3) + \mathbf{k}(-10 + 9) \]
\[ = \mathbf{i}(-1) - \mathbf{j}(1) + \mathbf{k}(-1) = (-1, -1, -1) \]

### Step 3: Calculate the vector between points on the lines
Now, we find \(\mathbf{A_2} - \mathbf{A_1}\):
\[ \mathbf{A_2} - \mathbf{A_1} = (2, 1, -2) - (0, 0, 0) = (2, 1, -2) \]

### Step 4: Calculate the distance using the formula
The formula for the shortest distance \(d\) between two skew lines is given by:
\[ d = \frac{|\mathbf{b_1} \times \mathbf{b_2} \cdot (\mathbf{A_2} - \mathbf{A_1})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]

Calculating the dot product:
\[ \mathbf{b_1} \times \mathbf{b_2} \cdot (\mathbf{A_2} - \mathbf{A_1}) = (-1, -1, -1) \cdot (2, 1, -2) = (-1)(2) + (-1)(1) + (-1)(-2) = -2 - 1 + 2 = -1 \]
Thus, the absolute value is:
\[ |\mathbf{b_1} \times \mathbf{b_2} \cdot (\mathbf{A_2} - \mathbf{A_1})| = |-1| = 1 \]

Next, we calculate the magnitude of \(\mathbf{b_1} \times \mathbf{b_2}\):
\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{(-1)^2 + (-1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 5: Substitute into the distance formula
Now substituting into the distance formula:
\[ d = \frac{1}{\sqrt{3}} \]

### Final Answer
The magnitude of the shortest distance between the two lines is:
\[ \boxed{\frac{1}{\sqrt{3}}} \]