If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,what is the angle between them?

The Cartesian equation of the line passing through the point $(-1, 3, -2)$ and perpendicular to the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}$ is

Let the image of the point $P(1, 2, 3)$ in the line $L : \frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3}$ be $Q$. Let $R(\alpha, \beta, \gamma)$ be a point that divides the line segment $PQ$ internally in the ratio $1:3$. Then the value of $22(\alpha+\beta+\gamma)$ is equal to

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

The point of intersection of the lines $\vec{r}=2 \vec{b}+t(6 \vec{c}-\vec{a})$ and $\vec{r}=\vec{a}+s(\vec{b}-3 \vec{c})$ is

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