If the lines $\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2}$ and $\frac{x-11}{5}=\frac{y-3}{3}=\frac{z-1}{1}$ are perpendicular,then $\lambda=$

Find the coordinates of the point where the line passing through $(5, 1, 6)$ and $(3, 4, 1)$ crosses the $ZX$-plane.

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 $d_1$ is the shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$,and $d_2$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$,then the value of $\frac{32 \sqrt{3} d_1}{d_2}$ is:

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

The point of intersection of the lines $\vec{r}=2 \vec{b}+t(6 \vec{c}-\vec{a})$ and $\vec{r}=\vec{a}+s(\vec{b}-3 \vec{c})$ is