Let $A \equiv (\lambda + 2, 1 - 2\lambda, \lambda + 2)$ and $B \equiv (2k + 1, k, k + 1)$ where $\lambda, k \in \mathbb{R}$. Then the minimum distance between $A$ and $B$ is -

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$,then the sum of all possible values of $\alpha$ is

The cosine of the angle included between the lines $\overline{r}=(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})+\lambda(\hat{\imath}-2 \hat{\jmath}-2 \hat{k})$ and $\overline{r}=(\hat{\imath}+\hat{\jmath}+3 \hat{k})+\mu(3 \hat{\imath}+2 \hat{\jmath}-6 \hat{k})$ where $\lambda, \mu \in R$ is

Let a straight line $L$ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines $\frac{x-1}{2} = \frac{y+1}{1} = \frac{z-3}{-2}$ and $\frac{x-3}{1} = \frac{y-2}{3} = \frac{z+2}{4}$. If the line $L$ intersects the $yz$-plane at the point $Q$,then the distance between the points $P$ and $Q$ is:

The length of the perpendicular from the point $(0, 2, 3)$ to the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$ is:

$P, Q, R$ and $S$ are four points with the position vectors $3i-4j+5k, 0i+0j+4k, -4i+5j+1k$ and $-3i+4j+3k$,respectively. Then,the line $PQ$ meets the line $RS$ at the point