The coordinates of the point where the line joining the points $(3, 5, -7)$ and $(-2, 1, 8)$ is intersected by the $YOZ$ plane are

If the shortest distance between the lines $r=(3 \hat{i}+4 \hat{j}-2 \hat{k})+t(-\hat{i}+2 \hat{j}+\hat{k})$ and $r=(\hat{i}-7 \hat{j}-2 \hat{k})+s(\hat{i}+3 \hat{j}+2 \hat{k})$ is equivalent to the projection of $P=-2 \hat{i}+11 \hat{j}$ on $Q$, then a possible vector $Q$ is

Show that the line passing through the points $(4,7,8)$ and $(2,3,4)$ is parallel to the line passing through the points $(-1,-2,1)$ and $(1,2,5)$.

The angle between the lines $\frac{x-2}{2} = \frac{y-3}{-2} = \frac{z-5}{1}$ and $\frac{x-2}{1} = \frac{y-3}{2} = \frac{z-5}{2}$ is $ . . . . . . $. (in $^{\circ}$)

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are at right angles,then $k =$

If the lines $\frac{x - 5}{5m + 2} = \frac{2 - y}{5} = \frac{1 - z}{-1}$ and $\frac{x}{1} = \frac{2y + 1}{4m} = \frac{1 - z}{-3}$ are perpendicular to each other, then the value of $m$ is ...