1. Key Concepts and Formulas

   • Vector Equation of a Line: A line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is represented as $\vec{r} = \vec{a} + t\vec{b}$, where $t$ is a scalar parameter.
   • Shortest Distance Between Skew Lines: For two skew lines (non-parallel and non-intersecting) given by $L_1: \vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $L_2: \vec{r} = \vec{a}_2 + \mu \vec{b}_2$, the shortest distance $D$ between them is given by the formula: $D = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$ Here, $(\vec{a}_2 - \vec{a}_1)$ is a vector connecting any point on $L_1$ to any point on $L_2$, and $(\vec{b}_1 \times \vec{b}_2)$ is a vector perpendicular to both $L_1$ and $L_2$ (the common normal). The formula essentially finds the projection of the connecting vector onto the direction of the common normal.
   • Greatest Common Divisor (GCD): For two integers $m$ and $n$, $\operatorname{gcd}(m, n)=1$ means that $m$ and $n$ are coprime, i.e., they share no common prime factors other than 1.

2. Step-by-Step Solution

   Step 1: Identify Position and Direction Vectors for Each Line. First, we need to express both lines in the standard vector form $\vec{r} = \vec{a} + t\vec{b}$ to clearly identify the position vectors ($\vec{a}_1, \vec{a}_2$) and direction vectors ($\vec{b}_1, \vec{b}_2$).

   For line $L_1$: $\vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}$ We separate the terms independent of $\lambda$ and the terms multiplied by $\lambda$: $\vec{r} = (2\hat{i} + \hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 4\hat{k})$ Thus, for $L_1$: $\vec{a}_1 = 2\hat{i} + \hat{j} + 3\hat{k}$ (Position vector of a point on $L_1$) $\vec{b}_1 = \hat{i} - 3\hat{j} + 4\hat{k}$ (Direction vector of $L_1$)

   For line $L_2$: $\vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}$ First, distribute the constants: $\vec{r} = (2+2\mu) \hat{i} + (3+3\mu) \hat{j} + (5+\mu) \hat{k}$ Now, separate the terms independent of $\mu$ and the terms multiplied by $\mu$: $\vec{r} = (2\hat{i} + 3\hat{j} + 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$ Thus, for $L_2$: $\vec{a}_2 = 2\hat{i} + 3\hat{j} + 5\hat{k}$ (Position vector of a point on $L_2$) $\vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k}$ (Direction vector of $L_2$)

   Step 2: Calculate the Vector Connecting Points on the Lines ($\vec{a}_2 - \vec{a}_1$). This vector represents the displacement from a point on $L_1$ to a point on $L_2$. $\vec{a}_2 - \vec{a}_1 = (2\hat{i} + 3\hat{j} + 5\hat{k}) - (2\hat{i} + \hat{j} + 3\hat{k})$ $\vec{a}_2 - \vec{a}_1 = (2-2)\hat{i} + (3-1)\hat{j} + (5-3)\hat{k}$ $\vec{a}_2 - \vec{a}_1 = 0\hat{i} + 2\hat{j} + 2\hat{k}$

   Step 3: Calculate the Cross Product of Direction Vectors ($\vec{b}_1 \times \vec{b}_2$). The cross product of the direction vectors gives a vector that is perpendicular to both lines $L_1$ and $L_2$. This vector defines the direction of the shortest distance between the two lines. $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 4 \\ 2 & 3 & 1 \end{vmatrix}$ Expanding the determinant: $= \hat{i}((-3)(1) - (4)(3)) - \hat{j}((1)(1) - (4)(2)) + \hat{k}((1)(3) - (-3)(2))$ $= \hat{i}(-3 - 12) - \hat{j}(1 - 8) + \hat{k}(3 + 6)$ $= -15\hat{i} - (-7)\hat{j} + 9\hat{k}$ $\vec{b}_1 \times \vec{b}_2 = -15\hat{i} + 7\hat{j} + 9\hat{k}$

   Step 4: Calculate the Magnitude of the Cross Product ($|\vec{b}_1 \times \vec{b}_2|$). This magnitude is needed to normalize the common normal vector in the shortest distance formula. $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-15)^2 + 7^2 + 9^2}$ $= \sqrt{225 + 49 + 81}$ $= \sqrt{355}$

   Step 5: Calculate the Scalar Triple Product ($(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$). This is the numerator of our shortest distance formula. It represents the scalar projection of the vector connecting the two lines onto their common normal. $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (0\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (-15\hat{i} + 7\hat{j} + 9\hat{k})$ $= (0)(-15) + (2)(7) + (2)(9)$ $= 0 + 14 + 18$ $= 32$

   Step 6: Apply the Shortest Distance Formula. Now, substitute all the calculated values into the shortest distance formula: $D = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$ $D = \left| \frac{32}{\sqrt{355}} \right|$ Since $32$ and $\sqrt{355}$ are positive, the absolute value is simply the expression itself: $D = \frac{32}{\sqrt{355}}$

   Step 7: Determine $m$ and $n$ and Verify $\operatorname{gcd}(m, n)=1$. The problem states the shortest distance is $\frac{m}{\sqrt{n}}$, where $\operatorname{gcd}(m, n)=1$. Comparing our result $D = \frac{32}{\sqrt{355}}$ with the given form, we identify: $m = 32$ $n = 355$

   Next, we verify the condition $\operatorname{gcd}(m, n)=1$. Prime factorization of $m = 32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5$. Prime factorization of $n = 355$. We can test for small prime factors: $355$ is divisible by $5$ (since it ends in $5$). $355 = 5 \times 71$. Both $5$ and $71$ are prime numbers. The prime factors of $m$ are $\{2\}$. The prime factors of $n$ are $\{5, 71\}$. Since there are no common prime factors, $\operatorname{gcd}(32, 355) = 1$. The condition is satisfied.

   Step 8: Calculate $m+n$. Finally, we calculate the required sum: $m+n = 32 + 355$ $m+n = 387$

3. Common Mistakes & Tips

   • Incorrect Vector Extraction: A common error is misidentifying $\vec{a}$ and $\vec{b}$ from the line equations, especially if the parameters are inside parentheses (as in $L_2$) or if signs are misinterpreted. Always expand and group terms carefully.
   • Cross Product Errors: Calculation of the determinant for the cross product can be prone to arithmetic mistakes or sign errors. Double-check each component.
   • Forgetting Absolute Value: The shortest distance must always be a non-negative value. Remember the absolute value in the formula.
   • Neglecting GCD Condition: If a GCD condition is specified, it's crucial to verify it. Sometimes, simplification might be needed (e.g., if the distance was $\frac{64}{\sqrt{1420}}$, it would need to be simplified to $\frac{32}{\sqrt{355}}$ before identifying $m$ and $n$).

4. Summary

   To find the shortest distance between the two skew lines, we first extracted their position and direction vectors. Then, we calculated the vector connecting points on the lines ($\vec{a}_2 - \vec{a}_1$) and the common normal vector ($\vec{b}_1 \times \vec{b}_2$). After finding the magnitude of the common normal and the scalar triple product, we applied the shortest distance formula. The calculated distance was $\frac{32}{\sqrt{355}}$. By comparing this to the given form $\frac{m}{\sqrt{n}}$ and verifying the $\operatorname{gcd}(m, n)=1$ condition, we found $m=32$ and $n=355$. Finally, their sum $m+n$ was calculated to be $387$.

The final answer is $\boxed{387}$, which corresponds to option (A).