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

Step 1: Identify points and direction vectors

First, we identify the position vectors of points on each line and their respective direction vectors. For the first line, a point on the line is $A_1(2, 1, -3)$ and its direction vector is $\vec{b}_{1} = \hat{i} + 2\hat{j} + 3\hat{k}$. For the second line, a point on the line is $A_2(-1, -3, -5)$ and its direction vector is $\vec{b}_{2} = 2\hat{i} + 4\hat{j} + 5\hat{k}$.

Step 2: Calculate vector connecting points

Next, we calculate the vector connecting a point on the first line to a point on the second line, which is $\vec{a}_{2} - \vec{a}_{1}$. This vector is obtained by subtracting the coordinates of $A_1$ from $A_2$.

Step 3: Calculate cross product of direction vectors

We then compute the cross product of the direction vectors $\vec{b}_{1}$ and $\vec{b}_{2}$. This cross product gives a vector that is perpendicular to both direction vectors, which is essential for finding the shortest distance.

Step 4: Calculate shortest distance

The shortest distance $d$ between two skew lines is given by the formula involving the scalar triple product. We substitute the calculated vectors into the formula and simplify to find the distance.

Step 5: Calculate the square of the shortest distance and m+n

The problem asks for the square of the shortest distance. We square the calculated distance to get $\frac{4}{5}$. Here, $m=4$ and $n=5$. Since 4 and 5 are coprime, we find $m+n = 4+5=9$.

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