The distance of the point $P(1, 2, 1)$ from the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-3}{2}$ is

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?

The shortest distance between the lines $\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}$ and $\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$ is $......$.

Let the line of the shortest distance between the lines $L_1: \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $L_2: \vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})$ intersect $L_1$ and $L_2$ at $P$ and $Q$ respectively. If $(\alpha, \beta, \gamma)$ is the midpoint of the line segment $PQ$,then $2(\alpha+\beta+\gamma)$ is equal to . . . . . . .

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 $k=$

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: