The shortest distance between lines $L_1:\frac{x+3}{-4}=\frac{6-y}{-3}=\frac{z}{2}$ and $L_2:\frac{x+2}{-4}=\frac{y}{1}=\frac{z-7}{1}$ is:

The shortest distance between the lines whose vector equations are $\vec{r} = (6\hat{i}+2\hat{j}+2\hat{k}) + \lambda(\hat{i}-2\hat{j}+2\hat{k})$ and $\vec{r} = (-4\hat{i}-\hat{k}) + \mu(3\hat{i}-2\hat{j}-2\hat{k})$

Let $L_1$ and $L_2$ lines such that $L_1:\vec{r}=\hat{i}+\hat{j}+\lambda(2\hat{i}-\hat{j}+\hat{k})$, $L_2:\vec{r}=2\hat{i}+\hat{j}-\hat{k}+\mu(4\hat{i}-2\hat{j}+2\hat{k})$ and $\theta$ be the acute angle between $L_1$ and $L_2$, then

Match the LIST-I with LIST-II

Choose the correct answer from the options given below:

If the line $\frac{-x+1}{3} = \frac{-y-2}{-2k} = \frac{z+3}{2}$ and $\frac{-1+x}{3k} = \frac{-1 +y}{1} = \frac{-z+6}{5}$ are perpendicular, then the value of k is:

The points on the line $\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ at a distance of $\sqrt{17}$ from the point (-2, -1, 3) are:

If a line makes angles $\alpha$, $\beta$ and $\gamma$ with the positive directions of coordinate axes respectively, then $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$ is equal to

The vector equation of the line joining the points $(-2, -3, -4)$ and $(1, -2, 4)$ is :