If the shortest distance between the lines $\vec r=(\hat i+2\hat j+3\hat k)+\lambda(2\hat i+3\hat j+4\hat k)$ and $\vec r=(a\hat i+4\hat j+5\hat k)+\mu(4\hat i+6\hat j+8\hat k)$, $a \in N$, where $\lambda$ and $\mu$ are parameters, is $\sqrt{\dfrac{5}{29}}$, then $a$ is equal to

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

The equation of line passing through the points (2, -1, 3) and (4, 2, 1) is

(A) $\frac{x-2}{2} = \frac{y+1}{3} = \frac{z-3}{-2}$

(B) $\frac{x-4}{3} = \frac{y-2}{-2} = \frac{z-1}{2}$

(C) $\vec{r} = (2\hat{i}-\hat{j}+3\hat{k}) + \lambda(2\hat{i}+3\hat{j}-2\hat{k})$

(D) $\frac{x-4}{-2} = \frac{y-2}{-3} = \frac{z-1}{2}$

Choose the correct answer from the options given below:

The shortest distance between the lines $\vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu(4\hat{i} - 2\hat{j} + 2\hat{k})$ is

If the lines $\frac{1-x}{3} = \frac{3y-6}{k} = \frac{3-z}{-2}$ and $\frac{1-x}{2k} = \frac{y-5}{3} = \frac{6-z}{5}$ are perpendicular to each other, then $k$ is equal to

Consider the line $\frac{x-2}{2} = \frac{2y-5}{-3}, z = -1$. Then which of the following is/are true?

(A) It has Direction ratios (2, -3, -1)

(B) It has Direction cosines $\left(\frac{4}{5}, \frac{-3}{5}, \frac{-1}{5}\right)$

(C) It has Direction ratios $\left(2, \frac{-3}{2}, 0\right)$

(D) It has Direction cosines $\left(\frac{4}{5}, \frac{-3}{5}, 0\right)$

Choose the correct answer from the options given below:

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: