The shortest distance between the lines $\frac{x-4}{1}=\frac{y-3}{2}=\frac{z-2}{-3}$ and $\frac{x+2}{2}=\frac{y-6}{4}=\frac{z-5}{-5}$ is:

Let the image of the point $\mathrm{P}(1,6, a)$ in the line $\mathrm{L}: \frac{x}{1}=\frac{y-1}{2}=\frac{z-a+1}{b}, b>0$, be $\left(\frac{a}{3}, 0, a+c\right)$. If $\mathrm{S}(\alpha, \beta, \gamma), \alpha>0$, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is $2 \sqrt{14}$, then $\alpha+\beta+\gamma$ is equal to:

Let a line L be perpendicular to both the lines $\mathrm{L}_1: \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and $\mathrm{L}_2: \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}$.

If $\theta$ is the acute angle between the lines L and $\mathrm{L}_3: \frac{x-\frac{8}{7}}{2}=\frac{y-\frac{4}{7}}{1}=\frac{z}{2}$, then $\tan \theta$ is equal to:

Let a triangle PQR be such that P and Q lie on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ and are at a distance of 6 units from $R(1,2,3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\Delta P Q R$, then $\alpha+\beta+\gamma$ is equal to :