Find the angle between the lines $\frac{x-2}{3} = \frac{y+1}{-2}, z=2$ and $\frac{x-1}{1} = \frac{2y+3}{3} = \frac{z+5}{2}$.

If $(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from a point $P(-1, 2, -1)$ to the line joining the points $A(2, -1, 1)$ and $B(1, 1, -2)$,then $\alpha + \beta + \gamma =$

Let the line passing through the points $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ intersect the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4,-5,1)$ is:

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

The equation of the line passing through the points $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ is given by:

The lines $\overrightarrow{r} = (\hat{i} - \hat{j}) + \ell(2\hat{i} + \hat{k})$ and $\overrightarrow{r} = (2\hat{i} - \hat{j}) + m(\hat{i} + \hat{j} - \hat{k})$: