Find the magnitude of the shortest distance between the lines \(\frac{x-0}{2} = \frac{y-0}{-3}=\frac {z-0}{1} \) and \(\frac{x-2}{3} = \frac{y-1}{-5}=\frac {z+2}{2} \).

Concept: 

The magnitude of the shortest distance between the lines \( \vec{r_1} = \vec a_1 + \lambda \vec b_1\) and \(\vec{r_2} = \vec a_2 + \mu\vec b_2\) is 

\(d = \left | \frac{(\vec b_1\times\vec b_2).(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|} \right|\)

Given:  

The lines \(\frac{x-0}{2} = \frac{y-0}{-3}=\frac {z-0}{1} \) and \(\frac{x-2}{3} = \frac{y-1}{-5}=\frac {z+2}{2} \).

Rewriting the given equations,

\( \vec{r_1} = \lambda(2\vec i-3\vec j+\vec k) \) and \( \vec{r_2} = (2\vec i+\vec j-2\vec k) + \mu(3\vec i-5\vec j+2\vec k) \)

⇒ \(\vec a_1=0\) ,  \(\vec b_1=2\vec i-3\vec j+\vec k\) and  \(\vec a_2=2\vec i+\vec j-2\vec k\),  \(\vec b_2=3\vec i-5\vec j+2\vec k\)

Therefore, the magnitude of the shortest distance between the given lines is

\(d = \left | \frac{(\vec b_1\times\vec b_2).(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|} \right|\)

\(d = \left | \frac{[(2\vec i-3\vec j+\vec k)\times(3\vec i-5\vec j+2\vec k)].[(2\vec i+\vec j-2\vec k)-0]}{|(2\vec i-3\vec j+\vec k)\times(3\vec i-5\vec j+2\vec k)|} \right|\)

\(d = \left | \frac{(-\vec i-\vec j-\vec k).(2\vec i+\vec j-2\vec k)]}{|-\vec i-\vec j-\vec k|} \right|\)

\(d = \frac{1}{\sqrt{3}}\)

Therefore, the magnitude of the shortest distance between the given lines is \(\frac{1}{\sqrt3}\).