If the shortest distance between the lines $\vec{r}_{1}=\alpha \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k}), \lambda \in R, \alpha>0$ and $\vec{r}_{2}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k}), \mu \in R$ is $9$,then $\alpha$ is equal to $.....$

The line $L$ given by $\frac{x - 2}{2} = \frac{y - 1}{b} = \frac{z + 1}{c}$ passes through the point $(1, 2, 3)$. Another line $K$ is parallel to line $L$ and has the equation $\frac{x + 2}{a} = \frac{y - 3}{2} = \frac{z + 4}{d}$. Then the distance between line $L$ and $K$ is

The foot of the perpendicular from $P(5, 7, 3)$ to the line joining $A(9, 13, 15)$ and $B(12, 21, 10)$ is:

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.

The equation of the line passing through the point $(1, -3, 5)$ and making equal angles with the coordinate axes is:

The Cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{6-z}{2}$. Find the vector equation of the line.