The shortest distance between the lines $r = (3t - 4)\hat{i} - 2\hat{j} - (1 + 2t)\hat{k}$ and $r = (6 + s)\hat{i} + (2 - 2s)\hat{j} + 2(1 + s)\hat{k}$ is

If $2x - y + z = 0 = y - x + 2z = mx - 2y + mz$ represents a line in space,then the value of $m$ is-

Find the angle between the lines $\vec{r}=3 \hat{i}-2 \hat{j}+6 \hat{k}+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})$ and $\vec{r}=(2 \hat{j}-5 \hat{k})+\mu(6 \hat{i}+3 \hat{j}+2 \hat{k})$.

The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is

The line $L$ given by $\frac{x - 2}{2} = \frac{y - 1}{b} = \frac{z + 1}{c}$ passes through the point $(1, 2, 3)$. Another line $K$ is parallel to line $L$ and has the equation $\frac{x + 2}{a} = \frac{y - 3}{2} = \frac{z + 4}{d}$. Then the distance between line $L$ and $K$ is

$A$ line having direction ratios $1, -4, 2$ intersects the lines $\frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1}$ and $\frac{x}{2} = \frac{y-7}{3} = \frac{z}{1}$ at the points $A$ and $B$ respectively. Then,the coordinates of points $A$ and $B$ are: