The position vectors of the vertices of a quadrilateral $ABCD$ are $a, b, c$ and $d$ respectively. The area of the quadrilateral formed by joining the midpoints of its sides is

The vector $\frac{1}{3}(2i - 2j + k)$ is

In a quadrilateral $ABCD$, the point $P$ divides $DC$ in the ratio $1:2$ and $Q$ is the midpoint of $AC$. If $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}=k\overrightarrow{PQ}$, then $k$ is equal to

Let $\vec{a}_n = (\tan \theta_n)\hat{i} + \hat{j}$ and $\vec{b}_n = \hat{i} - (\cot \theta_n)\hat{j}$, where $\theta_n = \frac{2^{n-1}\pi}{2^n+1}$, for some $n \in N, n > 5$. Then the value of $\frac{\sum_{k=1}^n |\vec{a}_k|^2}{\sum_{k=1}^n |\vec{b}_k|^2}$ is . . . . . . .

Match the statements given in Column $I$ with the values given in Column $II$.

Column $I$	Column $II$
$(A)$ If $\vec{a}=\hat{j}+\sqrt{3} \hat{k}, \vec{b}=-\hat{j}+\sqrt{3} \hat{k}$ and $\vec{c}=2 \sqrt{3} \hat{k}$ form a triangle,then the internal angle of the triangle between $\vec{a}$ and $\vec{b}$ is	$(p)$ $\frac{\pi}{6}$
$(B)$ If $\int_a^b(f(x)-3 x) d x=a^2-b^2$,then the value of $f\left(\frac{\pi}{6}\right)$ is	$(q)$ $\frac{2 \pi}{3}$
$(C)$ The value of $\frac{\pi^2}{\ln 3} \int_{1 / 6}^{5 / 6} \sec (\pi x) d x$ is	$(r)$ $\frac{\pi}{3}$
$(D)$ The maximum value of $|\operatorname{Arg}(\frac{1}{1-z})|$ for $|z|=1, z \neq 1$ is given by	$(s)$ $\pi$
	$(t)$ $\frac{\pi}{2}$

Let $\hat{i}, \hat{j}$ and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let $\vec{a}=3\hat{i}+\hat{j}-\hat{k}$,$\vec{b}=\hat{i}+b_2\hat{j}+b_3\hat{k}$ $(b_2, b_3 \in \mathbb{R})$,and $\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$ $(c_1, c_2, c_3 \in \mathbb{R})$ be three vectors such that $b_2b_3 > 0$,$\vec{a} \cdot \vec{b} = 0$ and $\begin{bmatrix} 0 & -c_3 & c_2 \\ c_3 & 0 & -c_1 \\ -c_2 & c_1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ b_2 \\ b_3 \end{bmatrix} = \begin{bmatrix} 3-c_1 \\ 1-c_2 \\ -1-c_3 \end{bmatrix}$. Then,which of the following is/are $TRUE$?