Two adjacent sides of a parallelogram ABCD are given by $\overline{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\overline{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side AD is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that AD becomes AD'. If AD' makes a right angle with the side AB, then the cosine of the angle $\alpha$ is given by

If $\theta$ is the angle between the lines $\dfrac{x-1}{2} = \dfrac{2y+3}{4};\ z = -2$ and $x = 1;\ \dfrac{y-1}{2} = \dfrac{z+1}{2}$, then

If for some $m \in \mathbb{R}$ the lines $L_1 : \dfrac{x+1}{m} = \dfrac{y-m}{-1} = \dfrac{z-1}{1}$ and $L_2 : \dfrac{x+2}{-4} = \dfrac{y+1}{9} = \dfrac{z+1}{1}$ are coplanar, then line $L_1$ passes through the point

If the lines $\vec{r} = \hat{i} + \hat{j} - \hat{k} + \lambda(q\hat{i} - 2\hat{j} + \hat{k})$ and $\vec{r} = p\hat{i} - 3\hat{j} + 2\hat{k} + \mu(\hat{i} - 2\hat{j} + 2\hat{k})$ intersect each other and $q\hat{i} - 2\hat{j} + \hat{k}$ is collinear to $4\hat{i} - 4\hat{j} + 2\hat{k}$, then the values of $p$ and $q$ are