The equation of the straight line $3x + 2y - z - 4 = 0$ and $4x + y - 2z + 3 = 0$ in the symmetrical form is:

The shortest distance between the line passing through the point $\bar{i} + 2\bar{j} + 3\bar{k}$ and parallel to the vector $2\bar{i} + 3\bar{j} + 4\bar{k}$ and the line passing through the point $2\bar{i} + 4\bar{j} + 5\bar{k}$ and parallel to the vector $3\bar{i} + 4\bar{j} + 5\bar{k}$ is:

Let $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?

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 shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$ is (in $\sqrt{3}$)

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