If $\overline{\mathrm{a}}=\frac{1}{\sqrt{10}}(3 \hat{\mathrm{i}}+\hat{\mathrm{k}}), \overline{\mathrm{b}}=\frac{1}{7}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})$, then the value of $(\overline{\mathrm{a}}-2 \overline{\mathrm{~b}}) \cdot\{(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(2 \overline{\mathrm{a}}+\overline{\mathrm{b}})\}$ is

The vectors $\bar{p}=\hat{i}+a \hat{j}+a^2 \hat{k}, \bar{q}=\hat{i}+b \hat{j}+b^2 \hat{k}$ and $\overline{\mathrm{r}}=\hat{\mathrm{i}}+\mathrm{c} \hat{\mathrm{j}}+\mathrm{c}^2 \hat{\mathrm{k}}$ are non-coplanar and $\left|\begin{array}{lll}a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3\end{array}\right|=0$ then the value of $(a b c)$ is

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

The general solutions of the equation $\tan ^2 \theta+\sec 2 \theta=1$ are