If the shortest distance between the lines $$\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$$ and $$\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$$ is $$\frac{13}{\sqrt{29}}$$, then a value of $$\lambda$$ is :

Let $$P(x, y, z)$$ be a point in the first octant, whose projection in the $$x y$$-plane is the point $$Q$$. Let $$O P=\gamma$$; the angle between $$O Q$$ and the positive $$x$$-axis be $$\theta$$; and the angle between $$O P$$ and the positive $$z$$-axis be $$\phi$$, where $$O$$ is the origin. Then the distance of $$P$$ from the $$x$$-axis is

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 :