The equation of the line passing through the points $(3, 4, -7)$ and $(6, -1, 1)$ is

If $p$ is the shortest distance between the lines $\frac{x + 1}{7} = \frac{y + 1}{-6} = \frac{z + 1}{1}$ and $\vec{r} = (3\hat{i} + 5\hat{j} + 7\hat{k}) + \mu(\hat{i} - 2\hat{j} + \hat{k})$, then $[p]$ is... (where $[.]$ denotes the greatest integer function.)

Line $L_1$ passes through the point $(1, 2, 3)$ and is parallel to the $z$-axis. Line $L_2$ passes through the point $(\lambda, 5, 6)$ and is parallel to the $y$-axis. Let for $\lambda = \lambda_1, \lambda_2$ with $\lambda_2 < \lambda_1$,the shortest distance between the two lines be $3$. Then the square of the distance of the point $(\lambda_1, \lambda_2, 7)$ from the line $L_1$ is:

$A$ line with direction ratios $2, 1, 2$ meets the lines $x = y + 2 = z$ and $x + 2 = 2y = 2z$ at points $P$ and $Q$ respectively. If the length of the perpendicular from the point $(1, 2, 12)$ to the line $PQ$ is $l$,then $l^2$ is

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect,then $k =$

If the two lines $l_{1}: \frac{x-2}{3}=\frac{y+1}{-2}, z=2$ and $l_{2}: \frac{x-1}{1}=\frac{2y+3}{\alpha}=\frac{z+5}{2}$ are perpendicular,then the angle between the lines $l_{2}$ and $l_{3}: \frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}$ is