The foot of the perpendicular drawn from the point $(1, 8, 4)$ on the line joining the points $(0, -11, 4)$ and $(2, -3, 1)$ is

The vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$ is . . . . . . .

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 $k = . . . . .$.

If $P$ is a point on the line parallel to the vector $2 \hat{i}-3 \hat{j}-6 \hat{k}$ and passing through the point $A$ whose position vector is $\hat{i}+2 \hat{j}-2 \hat{k}$ and $AP=21$, then the position vector of $P$ can be

The point of intersection of the line joining the points $(3, 4, 1)$ and $(5, 1, 6)$ and the $xy$-plane is

If the line joining $(2,3,-1)$ and $(3,5,-3)$ is perpendicular to the line joining $A(1,2,3)$ and $B(\alpha, \beta, \gamma)$,then a possible point for $B$ is