The straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is

The angle between the pair of lines $\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}$ and $\frac{x+1}{1}=\frac{y-4}{4}=\frac{z-5}{2}$ is

The shortest distance between the lines $\overline{r}=(3 \bar{i}-5 \bar{j}+2 \bar{k})+t(4 \bar{i}+3 \bar{j}-\bar{k})$ and $\overline{r}=(\bar{i}+2 \bar{j}-4 \bar{k})+s(6 \bar{i}+3 \bar{j}-2 \bar{k})$ is

The direction cosines of two lines are connected by the relations $l+m-n=0$ and $lm-2mn+nl=0$. If $\theta$ is the acute angle between those lines,then $\cos \theta=$

Line $L_1$ passes through the point $(1, 2, 3)$ and is parallel to the $z$-axis. Line $L_2$ passes through the point $(\lambda, 5, 6)$ and is parallel to the $y$-axis. Let for $\lambda = \lambda_1, \lambda_2$ with $\lambda_2 < \lambda_1$,the shortest distance between the two lines be $3$. Then the square of the distance of the point $(\lambda_1, \lambda_2, 7)$ from the line $L_1$ is:

The distance of the line $3y - 2z - 1 = 0 = 3x - z + 4$ from the point $(2, -1, 6)$ is: