Let ABCD and AEFG be squares of side 4 and 2 units, respectively. The point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies :

Let $$(\alpha, \beta, \gamma)$$ be the image of the point $$(8,5,7)$$ in the line $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$$. Then $$\alpha+\beta+\gamma$$ is equal to :

Let $$S_1=\{z \in \mathbf{C}:|z| \leq 5\}, S_2=\left\{z \in \mathbf{C}: \operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0\right\}$$ and $$S_3=\{z \in \mathbf{C}: \operatorname{Re}(z) \geq 0\}$$. Then the area of the region $$S_1 \cap S_2 \cap S_3$$ is :

If the constant term in the expansion of $$\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$$, is $$\alpha \times 2^8 \times \sqrt[5]{3}$$, then $$25 \alpha$$ is equal to :