Find the S.D. between the lines :
`(i) (x)/(2) = (y)/(-3) = (z)/(1) and (x -2)/(3) = (y - 1)/(-5) = (z + 4)/(2)`
(ii) `(x -1)/(2) = (y - 2)/(3) = (z - 3)/(2) and (x + 1)/(3) = (y - 1)/(2) = (z - 1)/(5)`
(iii) ` (x + 1)/(7) = (y + 1)/(-6) = (z + 1)/(1) and (x -3)/(1) = (y -5)/(-2) = (z - 7)/(1)`
(iv) `(x - 3)/(3) = (y - 8)/(-1) = (z-3)/(1) and (x + 3)/(-3) = (y +7)/(2) = (z -6)/(4) ` .

Answer

To find the shortest distance between two lines in three-dimensional geometry, we can use the formula:

\[ D_{min} = \frac{|(C - A) \cdot (B \times D)|}{|B \times D|} \]

where:
- \(A\) and \(C\) are points on the first and second lines, respectively.
- \(B\) and \(D\) are direction vectors of the first and second lines, respectively.
- \(\cdot\) denotes the dot product.
- \(\times\) denotes the cross product.

### Step-by-step Solution:

#### (i) For the lines:
1. **Line 1**: \(\frac{x}{2} = \frac{y}{-3} = \frac{z}{1}\)
- Point \(A(0, 0, 0)\) (when \(t = 0\))
- Direction vector \(B = (2, -3, 1)\)

2. **Line 2**: \(\frac{x - 2}{3} = \frac{y - 1}{-5} = \frac{z + 4}{2}\)
- Point \(C(2, 1, -4)\) (when \(s = 0\))
- Direction vector \(D = (3, -5, 2)\)

3. **Calculate \(C - A\)**:
\[ C - A = (2 - 0, 1 - 0, -4 - 0) = (2, 1, -4) \]

4. **Calculate \(B \times D\)**:
\[ B \times D = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 3 & -5 & 2 \end{vmatrix} = \hat{i}((-3)(2) - (1)(-5)) - \hat{j}((2)(2) - (1)(3)) + \hat{k}((2)(-5) - (-3)(3)) \]
\[ = \hat{i}(-6 + 5) - \hat{j}(4 - 3) + \hat{k}(-10 + 9) = \hat{i}(-1) - \hat{j}(1) + \hat{k}(-1) = (-1, -1, -1) \]

5. **Calculate \(|B \times D|\)**:
\[ |B \times D| = \sqrt{(-1)^2 + (-1)^2 + (-1)^2} = \sqrt{3} \]

6. **Calculate \((C - A) \cdot (B \times D)\)**:
\[ (C - A) \cdot (B \times D) = (2, 1, -4) \cdot (-1, -1, -1) = 2(-1) + 1(-1) + (-4)(-1) = -2 - 1 + 4 = 1 \]

7. **Calculate \(D_{min}\)**:
\[ D_{min} = \frac{|1|}{\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \]