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

Let $$\mathrm{P}(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{Q}(3,-3,1)$$ in the line $$\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$$ and $$\mathrm{R}$$ be the point $$(2,5,-1)$$. If the area of the triangle $$\mathrm{PQR}$$ is $$\lambda$$ and $$\lambda^2=14 \mathrm{~K}$$, then $$\mathrm{K}$$ is equal to :

If $$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1)$$ and $$D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$$ are the vertices of a quadrilateral $$A B C D$$, then its area is

The shortest distance between the lines $$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$$ and $$\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$$ is