If the lines $x = ay - 1 = z - 2$ and $x = 3y - 2 = bz - 2$ $(ab \neq 0)$ are coplanar,then

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 $k$ is equal to:

If $\bar{r} \cdot(2 \bar{i}+3 \bar{j}+4 \bar{k})=5$ and $\bar{r} \cdot(\bar{i}+\bar{j}-\bar{k})=7$ are two planes and $(16, -9, 0)$ is a point common to both the planes,then the vector equation of the line of intersection of the planes is $\bar{r}=$

Find the coordinates of the foot of the perpendicular drawn from the point $P(-1, 1, 2)$ to the plane $2x - 3y + z - 11 = 0$.

$A$ line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. Then the length of the line segment $PQ$ equals

Let the image of the point $P(1, 6, a)$ in the line $L: \frac{x-1}{2} = \frac{y-2}{b} = \frac{z-a+1}{1}, b>0$,be $Q(\frac{a}{3}, 0, a+c)$. If $S(\alpha, \beta, \gamma), \alpha > 0$,is the point on $L$ such that the distance of $S$ from the foot of perpendicular $F$ from the point $P$ on $L$ is $2\sqrt{14}$,then $\alpha + \beta + \gamma$ is equal to: