The angle between the lines $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$ is . . . . . . .

Let a line $L$ be perpendicular to both the lines $L_1: \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$ and $L_2: \frac{x-2}{1} = \frac{y-4}{4} = \frac{z-6}{7}$. If $\theta$ is the acute angle between the lines $L$ and $L_3: \frac{x-7}{2} = \frac{y-7}{1} = \frac{z}{2}$, then $\tan \theta$ is equal to:

Let $A \equiv (\lambda + 2, 1 - 2\lambda, \lambda + 2)$ and $B \equiv (2k + 1, k, k + 1)$ where $\lambda, k \in \mathbb{R}$. Then the minimum distance between $A$ and $B$ is -

Let $ABC$ be a triangle with $A(\alpha, 5, \beta)$, $B(-2, 1, 6)$ and $C(1, 0, -3)$ as its vertices. If the median through $B$ is equally inclined to the coordinate axes, then $\alpha + \beta =$

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

The coordinates of the foot of the perpendicular drawn from the point $2 \hat{i} - \hat{j} + 5 \hat{k}$ to the line $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k})$ are