What is the point of intersection of the line joining the points $(3, 4, 1)$ and $(5, 1, 6)$ with the $xy$-plane?

The distance between the parallel lines $\frac{x-1}{2}=\frac{y-2}{-2}=\frac{z-3}{1}$ and $\frac{x}{2}=\frac{y}{-2}=\frac{z}{1}$ is

Let $L_1$ (respectively $L_2$) be the line passing through $2 \hat{i}-\hat{k}$ (respectively $2 \hat{i}+\hat{j}-3 \hat{k}$) and parallel to $3 \hat{i}-\hat{j}+2 \hat{k}$ (respectively $\hat{i}-2 \hat{j}+\hat{k}$). Then the shortest distance between the lines $L_1$ and $L_2$ is equal to

If the shortest distance between the lines $\vec{r}=(-\hat{i}+3\hat{k})+\lambda(\hat{i}-a\hat{j})$ and $\vec{r}=(-\hat{j}+2\hat{k})+\mu(\hat{i}-\hat{j}+\hat{k})$ is $\sqrt{\frac{2}{3}}$,then the integral value of $a$ is equal to

If the shortest distance between the lines $r=(3 \hat{i}+4 \hat{j}-2 \hat{k})+t(-\hat{i}+2 \hat{j}+\hat{k})$ and $r=(\hat{i}-7 \hat{j}-2 \hat{k})+s(\hat{i}+3 \hat{j}+2 \hat{k})$ is equivalent to the projection of $P=-2 \hat{i}+11 \hat{j}$ on $Q$, then a possible vector $Q$ is

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