If the lines $\frac{x-k}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\frac{9}{2}}{2}=\frac{z}{1}$ intersect,then the value of $k$ is

If $P$ is a point lying on the line passing through the point $A(\hat{i}-\hat{j}+3 \hat{k})$ and parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$ such that $|AP|=18$,then a position vector of $P$ is

The equation of a line passing through the point $(2, 4, 6)$ and parallel to the line $3x + 4 = 4y - 1 = 1 - 4z$ is

The Cartesian equation of a line is $\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$. Find the vector equation for the line.

The angle between the lines $\vec{r}=(2 \hat{i}-3 \hat{j}+\hat{k})+\lambda(\hat{i}+4 \hat{j}+3 \hat{k})$ and $\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ is

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\lambda}{2}=\frac{z}{1}$ intersect each other,then $\lambda = \ldots$