The lines $\frac{x-3}{1}=\frac{y-2}{1}=\frac{z-5}{-k}$ and $\frac{x-4}{k}=\frac{y-3}{1}=\frac{z-3}{2}$ are coplanar,hence $k=$

Let $L_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $L_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in R$,be two lines,which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$,then the value of $26 \alpha(PB)^2$ is . . . . . . .

The vertices $B$ and $C$ of a triangle $ABC$ lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{z-2}{3}$. The coordinates of $A$ and $B$ are $(1, 6, 3)$ and $(4, 9, \alpha)$ respectively,and $C$ is at a distance of $10$ units from $B$. The area (in sq. units) of $\Delta ABC$ is:

The shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$ is (in $\sqrt{3}$)

If $A(3,-1,11)$,$B(0,2,3)$,and $C(4,8,11)$ are three points,then the coordinates of the foot of the perpendicular drawn from the point $A$ to the line joining the points $B$ and $C$ is

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