Interaction between Two Lines in 3D

If the lines  $\frac{x-1}{-3}=\frac{y-2}{-2k}=\frac{z-3}{2}$  and  $\frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}$ are perpendicular, what would be the value of $k$ and the equation of plane containing these lines?

Find the shortest distance between the lines $\overrightarrow{r}=3\hat{i}+2\hat{j}-4\hat{k}+\lambda (\hat{i}+2\hat{j}+2\hat{k})$and $\overrightarrow{r}=5\hat{i}-2\hat{j}+\mu (3\hat{i}+2\hat{j}+6\hat{k})$.

If the lines intersect find their point of intersection.

State whether following vectors are parallel or not.

$\overrightarrow{a}=3\hat{i}+2\hat{j}+5\hat{k} \& \overrightarrow{b}=6\hat{i}+8\hat{j}+10\hat{k}$

State whether following vectors are parallel or not.

$\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k} \& \overrightarrow{b}=3\hat{i}+6\hat{j}+9\hat{k}$

State whether following vectors are parallel or not.

$\overrightarrow{a}=2\hat{i}+4\hat{j}+5\hat{k} \& \overrightarrow{b}=6\hat{i}+8\hat{j}+10\hat{k}$

State whether following vectors are parallel or not.

$\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k} \& \overrightarrow{b}=2\hat{i}+4\hat{j}+6\hat{k}$

State whether following vectors are parallel or not.

$\overrightarrow{a}=2\hat{i}+4\hat{j}+5\hat{k} \& \overrightarrow{b}=4\hat{i}-8\hat{j}+10\hat{k}$

Let $l_0$ be the line defined by the vector (equation) $\hat{i}+2\hat{j}+3\hat{k}+\lambda (\hat{i}+\hat{j}+\hat{k})$, with $\lambda$ real. Which of the following vector equations, with $\mu$ real, defines a line which intersects $l_0$ ?

The value of $\alpha$ for which the shortest distance between the lines represented by $y+z=0, z+x=0$ and $x+y=0, x+y+z=\alpha$ is $1,$ is

The set of all non-zero real values of $k$, for which the lines $\frac{x-4}{2}=\frac{y-6}{2}=\frac{z-8}{-2k^2}$ and $\frac{x-2}{2k^2}=\frac{y-8}{4}=\frac{z-10}{2}$ are coplanar

If the shortest distance between the lines $x+2\lambda =2y=-12z, x=y+4\lambda =6z-12\lambda$ is $4\sqrt{2}$ units, then value of $\lambda$ is

If lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and  $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z-0}{1}$ intersect, then the value of $k$ is

The lines $\frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1}$ intersect each other at point