If the square of the shortest distance between the lines
$$ \frac{x-2}{1} = \frac{y-1}{2} = \frac{z+3}{-3} \quad \text{and} \quad \frac{x+1}{2} = \frac{y+3}{4} = \frac{z+5}{-5}, $$
is $ \frac{m}{n} $, where $m, n$ are coprime numbers, then $m+n$ is equal to:
The shortest distance between two skew lines is given by:
\[
d = \frac{|(\vec{a_1} - \vec{a_2}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|},
\]
where \( \vec{a_1}, \vec{a_2} \) are points on each line and \( \vec{b_1}, \vec{b_2} \) are the direction vectors of the lines.
Using the given data, calculate the shortest distance and square it to find \( \frac{m}{n} \). The result gives \( m + n = 9 \).
Thus, the correct answer is (b) 9.