If the square of the shortest distance between the lines $$\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}$$ and $$\frac{x+1}{2}=\frac{y+3}{4}=\frac{x+5}{-5}\text{ is }\frac{m}{n}$$, where m, n are coprime numbers, then m + n is equals to:

The first line is given by $$\frac{x-2}{1} = \frac{y-1}{2} = \frac{z+3}{-3}$$, with point $$(2, 1, -3)$$ and direction vector $$\vec{d_1}=(1,2,-3)$$

The second line is given by $$\frac{x+1}{2} = \frac{y+3}{4} = \frac{z+5}{-5}$$, with point $$(-1, -3, -5)$$ and direction vector $$\vec{d_2} = (2, 4, -5)$$.

Since $$(1,2,-3)$$ is not a scalar multiple of $$(2,4,-5)$$ , the lines are skew.

 $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix}  = \hat{i}(-10+12) - \hat{j}(-5+6) + \hat{k}(4-4) = 2\hat{i} - \hat{j} + 0\hat{k} = (2, -1, 0)$$.

The vector joining the given points on the two lines is $$\vec{P_1P_2} = (-1-2, -3-1, -5+3) = (-3, -4, -2)$$.

The shortest distance between skew lines is given by $$d = \frac{|\vec{P_1P_2} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}\,. $$ Here $$\vec{P_1P_2} \cdot (2, -1, 0) = (-3)(2) + (-4)(-1) + (-2)(0) = -6 + 4 + 0 = -2$$ and $$|\vec{d_1} \times \vec{d_2}| = \sqrt{4+1+0} = \sqrt{5}\,, $$ so $$d = \frac{|-2|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$$.

Squaring this distance yields $$d^2 = \frac{4}{5}\,, $$ so $$m = 4$$, $$n = 5$$ are coprime and $$m + n = 4 + 5 = 9\$$