The coordinates of the points on the line $\frac{x+2}{1}=\frac{y-1}{2}=\frac{z+1}{-2}$ at a distance of $12 \text{ units}$ from the point $A(-2, 1, -1)$ are

If the shortest distance between the lines $\overline{r}_1 = \alpha \hat{i} + 2 \hat{j} + 2 \hat{k} + \lambda(\hat{i} - 2 \hat{j} + 2 \hat{k})$ and $\overline{r}_2 = -4 \hat{i} - \hat{k} + \mu(3 \hat{i} - 2 \hat{j} - 2 \hat{k})$ is $9$,where $\lambda, \mu \in R$ and $\alpha > 0$,then the value of $\alpha$ is:

The equation of the line passing through the point $(0,1,2)$ and perpendicular to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is

If the Cartesian equation of the line is $x-1=2y+3=3-z$,then its vector equation is

The equation of the straight line $3x + 2y - z - 4 = 0$ and $4x + y - 2z + 3 = 0$ in the symmetrical form 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