Lines $\frac{x-5}{7}=\frac{y-5}{k}=\frac{z-2}{1}$ and $\frac{x}{1}=\frac{y-3}{2}=\frac{z+1}{3}$ are perpendicular to each other,then the value of $k=$ . . . . . . .

If the points $A(-1, 3, 2)$,$B(-4, 2, -2)$,and $C(5, 5, \lambda)$ are collinear,then $\lambda = $

The equation of the line passing through the point $(1, 2, 3)$ and perpendicular to the lines $\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3}$ and $\bar{r} = \lambda(-3 \hat{i} + 2 \hat{j} + 5 \hat{k})$ is

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 :

Find the position vector of the image of the point with position vector $\vec{P} = 2\hat{i} + \hat{j} + 3\hat{k}$ in the line whose vector equation is $\vec{r} = \hat{j} - 2\hat{k} + \lambda(\hat{i} + \hat{j} - \hat{k})$.

The angle between the lines $x=y, z=0$ and $y=0, z=0$ is (in $^{\circ}$)