The shortest distance (in units) between the lines $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda(\hat{i}+2\hat{j})$ is

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:

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

If lines $\frac{x - 1}{3} = \frac{y - 2}{-1} = \frac{z - \lambda}{2}$ and $\frac{x + 1}{-2} = \frac{y}{3\lambda} = \frac{2z - 7}{1}$ are coplanar,then the sum of the value$(s)$ of $\lambda$ is:

Point $A$ lies at a distance of $6$ units from the point $(1, 0, 1)$ on the line $\frac{x - 1}{2} = \frac{y}{2} = \frac{z - 1}{1}$ in the negative $z$-direction. Find the coordinates of $A$.

Statement $1:$ The shortest distance between the lines $\frac{x}{2} = \frac{y}{-1} = \frac{z}{2}$ and $\frac{x-1}{4} = \frac{y-1}{-2} = \frac{z-1}{4}$ is $\sqrt{2}$.
Statement $2:$ The shortest distance between two parallel lines is the perpendicular distance from any point on one of the lines to the other line.