The coordinates of the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ are

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 $\lambda=$

If the lines $\frac{x - 1}{c} = \frac{y + 2}{-2} = \frac{z - 3}{4}$ and $\frac{x - 5}{1} = \frac{y - 3}{1} = \frac{z + 1}{c}$ are parallel,then $c = ....$

If the shortest distance between the straight lines $3(x-1)=6(y-2)=2(z-1)$ and $4(x-2)=2(y-\lambda)=(z-3)$,$\lambda \in R$ is $\frac{1}{\sqrt{38}}$,then the integral value of $\lambda$ is equal to :

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is

$L_1$ is a line passing through the points with position vectors $\hat{i}-2 \hat{j}-\hat{k}$ and $4 \hat{i}-3 \hat{k}$. $L_2$ is a line passing through the points with position vectors $\hat{i}+2 \hat{j}-\hat{k}$ and $2 \hat{i}-4 \hat{j}-5 \hat{k}$. Then the distance between $L_1$ and $L_2$ is