The line $L$ passes through the point $(1, 2, 3)$. The distance of any point on the line $L$ from the line $\vec{r} = (-1, 3, 4) + \lambda(3, -2, 1)$ is constant. Then the line $L$ does not pass through the point:

Let $P(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z}{1}$ at a distance $4\sqrt{14}$ from the point $(1, -1, 0)$ and nearer to the origin. Then the shortest distance between the lines $\frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3}$ and $\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$ is equal to

The parametric equations of the line passing through $A(3, 4, -7)$ and $B(1, -1, 6)$ are:

If the two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ have a point in common,then $k=$

Statement-$1$: The shortest distance between the skew lines $\frac{x+3}{-4} = \frac{y-6}{3} = \frac{z}{2}$ and $\frac{x+3}{-4} = \frac{y}{1} = \frac{z-7}{1}$ is $9$.
Statement-$2$: Two lines are skew lines if there exists no plane passing through them.

The point having position vector $\vec{p} = 4\hat{i} - 11\hat{j} + 2\hat{k}$ lies on which of the following lines?