The area of the region in the first quadrant inside the circle $$x^2+y^2=8$$ and outside the parabola $$y^2=2 x$$ is equal to :

Let $$\overrightarrow{\mathrm{a}}=4 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=11 \hat{i}-\hat{j}+\hat{k}$$ and $$\overrightarrow{\mathrm{c}}$$ be a vector such that $$(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}})$$. If $$(2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670$$, then $$|\vec{c}|^2$$ is equal to:

If the line segment joining the points $$(5,2)$$ and $$(2, a)$$ subtends an angle $$\frac{\pi}{4}$$ at the origin, then the absolute value of the product of all possible values of $a$ is :

If $$\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$$ and $$\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$$, then $$\frac{\mathrm{a}}{\alpha-\mathrm{a}}+\frac{\mathrm{b}}{\beta-\mathrm{b}}+\frac{\gamma}{\gamma-\mathrm{c}}$$ is equal to :