Find the length of the perpendicular drawn from the point $P(3, -1, 11)$ to the line $\frac{x}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$.

If the line joining $A(4,1,2)$ and $B(0, k, 1)$ is perpendicular to the line joining $C(-2,1,1)$ and $D(4,2,5)$,then the value of $k$ is equal to

Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to ....................

Find the vector equation of the line $\frac{x - 2}{2} = \frac{2y - 5}{-3}, z = -1$.

The line $l_1$ passes through the point $(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. Then the shortest distance between the line $l_1$ and the line $\frac{x + 1}{2} = \frac{y + 4}{-3} = \frac{z}{2}$ is :

Let $L$ be a line passing through a point $A$ and parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$. Let $-7 \hat{i}-5 \hat{j}+11 \hat{k}$ be the position vector of a point $P$ on $L$ such that $|\overline{AP}|=12$. Then the position vector of $A$ can be