If the shortest distance between the lines

$$\begin{array}{ll} L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \\ L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, & \mu \in \mathbb{R} \end{array}$$

is $\frac{m}{\sqrt{n}}$, where $\operatorname{gcd}(m, n)=1$, then the value of $m+n$ equals

<p>$$\begin{aligned} & L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k} \\ & L_1=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+4 \hat{k}) \\ & L_2: \vec{r}=2 \hat{i}+3 \hat{j}+5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k}) \\ & \vec{a}_1=2 \hat{i}+\hat{j}+3 \hat{k} \\ & \vec{a}_2=2 \hat{i}+3 \hat{j}+5 \hat{k} \\ & \vec{a}_2-\vec{a}_1=2 \hat{j}+2 \hat{k} \\ & \vec{b}_1=\hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}_2=2 \hat{i}+3 \hat{j}+\hat{k} \\ & \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 4 \\ 2 & 3 & 1 \end{array}\right| \end{aligned}$$</p> <p>$$\begin{aligned} & \hat{i}(-3-12)-\hat{j}(1-8)+\hat{k}(3+6) \\ & =-15 \hat{i}+7 \hat{j}+9 \hat{k} \\ & \left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{225+49+81} \\ & \left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|=\frac{14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}} \\ & m+n=387 \end{aligned}$$</p>