If the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular,find the value of $k$.

The shortest distance between the lines $L_1: \bar{r} = \hat{i} + \hat{j} + \lambda(\hat{i} + \hat{j} - \hat{k})$ and $L_2: \bar{r} = \hat{j} + \hat{k} + \mu(\hat{j} + 2\hat{k} - \hat{i})$ is equal to:

Let $A$ and $B$ be two points on the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{-1}$. If the distance of point $P(1, 1, 1)$ from the points $A$ and $B$ is $\sqrt{3}$,then the distance between $A$ and $B$ is:

If the image of the point $P(a, 2, a)$ in the line $\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1}$ is $Q$ and the image of $Q$ in the line $\frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5}$ is $P$, then $a+b$ is equal to:

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 line joining the points $(1, 2, 3)$ and $(2, 3, 4)$,and the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}$ is $\alpha$,then $28 \alpha^2$ is equal to $........$.