Find the coordinates of the foot of the perpendicular drawn from the point $P(1, 0, 3)$ to the line joining the points $A(4, 7, 1)$ and $B(3, 5, 3)$.

If for some $\alpha \in R$,the lines $L_1: \frac{x+1}{2}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $L_2: \frac{x+2}{\alpha}=\frac{y+1}{5-\alpha}=\frac{z+1}{1}$ are coplanar,then the line $L_2$ passes through the point

If the shortest distance between the straight lines $3(x-1)=6(y-2)=2(z-1)$ and $4(x-2)=2(y-\lambda)=(z-3)$,$\lambda \in R$ is $\frac{1}{\sqrt{38}}$,then the integral value of $\lambda$ is equal to :

The equation of a line passing through the point $(-3, 2, -4)$ and equally inclined to the axes is:

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

Let $l_{1}$ be the line in $xy$-plane with $x$ and $y$ intercepts $\frac{1}{8}$ and $\frac{1}{4 \sqrt{2}}$ respectively,and $l_{2}$ be the line in $zx$-plane with $x$ and $z$ intercepts $-\frac{1}{8}$ and $-\frac{1}{6 \sqrt{3}}$ respectively. If $d$ is the shortest distance between the line $l_{1}$ and $l_{2}$,then $d^{-2}$ is equal to