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:

We need to find the square of the shortest distance between two lines.

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 $$(2,4,-5)$$ (one finds $$1/2 = 2/4$$ but $$-3/-5 = 3/5 \neq 1/2$$), the lines are skew.

We compute the cross product of the direction vectors by evaluating the determinant $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = \hat{i}(2 \times (-5) - (-3) \times 4) - \hat{j}(1 \times (-5) - (-3) \times 2) + \hat{k}(1 \times 4 - 2 \times 2) = \hat{i}(-10+12) - \hat{j}(-5+6) + \hat{k}(4-4) = 2\hat{i} - \hat{j} + 0\hat{k} = (2, -1, 0)\,$$.

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 $$m + n = 4 + 5 = 9\,$$.

The correct answer is Option 2: 9.