Let $\bar{a} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}),\ \bar{b} = (b_1\hat{i} + b_2\hat{j} + b_3\hat{k}),\ \bar{c} = (c_1\hat{i} + c_2\hat{j} + c_3\hat{k})$ be three non-zero vectors such that $\bar{a}$ is a unit vector perpendicular to both $\bar{b}$ and $\bar{c}$. If the angle between $\bar{b}$ and $\bar{c}$ is $\dfrac{\pi}{3}$ then $\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}^2 = $

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