If $|\overline{\mathrm{a}}|=2,|\overline{\mathrm{~b}}|=3$ and $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ are mutually perpendicular vectors, then the area of the triangle whose vertices are $0, a+2 b, a-2 b$ is

Let $\bar{A}, \bar{B}, \bar{C}$ be vectors of lengths 3 units, 4 units, 5 units respectively. let $\bar{A}$ be perpendicular to $\overline{\mathrm{B}}+\overline{\mathrm{C}}, \overline{\mathrm{B}}$ be perpendicular to $\overline{\mathrm{C}}+\overline{\mathrm{A}}$ and $\overline{\mathrm{C}}$ be perpendicular to $\bar{A}+\bar{B}$, then the length of vector $\overline{\mathrm{A}}+\overline{\mathrm{B}}+\overline{\mathrm{C}}$ is

Let $\bar{a}, \bar{b}$ and $\bar{c}$ be three non-zero vectors such that no two of them are collinear and $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{~b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}}$. If $\theta$ is the angle between vectors $\bar{b}$ and $\bar{c}$, then the value of $\operatorname{cosec} \theta$ is

Let $\quad \overline{\mathrm{a}}=\alpha \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \quad \overline{\mathrm{b}}=3 \hat{\mathrm{i}}-\beta \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \quad$ and $\overline{\mathrm{c}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$, where $\alpha, \beta \in \mathbb{R}$, be three vectors. If the projection at $\overline{\mathrm{a}}$ on $\overline{\mathrm{c}}$ is $\frac{10}{3}$ and $\overline{\mathrm{b}} \times \overline{\mathrm{c}}=-6 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$, then the value of $\alpha^2+\beta^2-\alpha \beta$ is equal to