Find the shortest distance between the lines given by $\vec{r}=(8+3 \lambda) \hat{i}+(-9-16 \lambda) \hat{j}+(10+7 \lambda) \hat{k}$ and $\vec{r}=15 \hat{i}+29 \hat{j}+5 \hat{k}+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k})$. (in $\text{ units}$)

Let $ABC$ be a triangle with $A(\alpha, 5, \beta)$,$B(-2, 1, 6)$ and $C(1, 0, -3)$ as its vertices. If the median through $B$ is equally inclined to the coordinate axes,then $\alpha + \beta =$

The lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are perpendicular to each other. Then,the value of $p$ is . . . . . . .

Let $(\alpha, \beta, \gamma)$ be the foot of the perpendicular from the point $(1, 2, 3)$ on the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. Then $19(\alpha + \beta + \gamma)$ is equal to:

Let $A(\alpha, 4, 7)$ and $B(3, \beta, 8)$ be two points in space. If the $YZ$ plane and $ZX$ plane respectively divide the line segment joining the points $A$ and $B$ in the ratio $2:3$ and $4:5$,then the point $C$ which divides $\overline{AB}$ in the ratio $\alpha: \beta$ externally is

The perpendicular distance of the point $(2, 4, -1)$ from the line $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9}$ is