If the shortest distance between the lines $\overline{r}_1 = \alpha \hat{i} + 2 \hat{j} + 2 \hat{k} + \lambda(\hat{i} - 2 \hat{j} + 2 \hat{k})$ and $\overline{r}_2 = -4 \hat{i} - \hat{k} + \mu(3 \hat{i} - 2 \hat{j} - 2 \hat{k})$ is $9$,where $\lambda, \mu \in R$ and $\alpha > 0$,then the value of $\alpha$ is:

$A(1, -2, 1)$ and $B(2, -1, 2)$ are the end points of a line segment. If $D(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from $C(1, 2, 3)$ to $AB$,then $\alpha^2 + \beta^2 + \gamma^2 =$

Given the lines $\vec{r} = (3+t)\hat{i} + (1-t)\hat{j} + (-2-2t)\hat{k}$,$t \in R$ and $x = 4+k, y = -k, z = -4-2k$,$k \in R$. What is the relationship between these two lines?

Let $P$ be the point of intersection of the lines $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ and $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$. Then,the shortest distance of $P$ from the line $4x=2y=z$ is

$A$ line $L_1$ passes through the point with position vector $3 \hat{i}$ and is parallel to the vector $-\hat{i}+\hat{j}+\hat{k}$. Another line $L_2$ passes through the point with position vector $\hat{i}+\hat{j}$ and is parallel to the vector $\hat{i}+\hat{k}$. Find the position vector of the point of intersection of lines $L_1$ and $L_2$.

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $\frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is