Considering only the principal values of an inverse function, the set

$$\mathrm{A}=\left\{x \geq 0 / \tan ^{-1} x+\tan ^{-1} 6 x=\frac{\pi}{4}\right\}$$

The line $$\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$$ lies in the plane $$x+3 y-\alpha z+\beta=0$$, then the value of $$\alpha^2+\alpha \beta+\beta^2$$ is

The vector projection of $$\overline{\mathrm{AB}}$$ on $$\overline{\mathrm{CD}}$$, where $$A \equiv(2,-3,0), B \equiv(1,-4,-2), C \equiv(4,6,8)$$ and $$\mathrm{D} \equiv(7,0,10)$$, is

If the circles $$x^2+y^2=9$$ and $$x^2+y^2+2 \alpha x+2 y+1=0$$ touch each other internally, then the value of $$\alpha^3$$ is