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

If $$y=\cos ^{-1}\left(\frac{\mathrm{a}^2}{\sqrt{x^4+\mathrm{a}^4}}\right)$$, then $$\frac{\mathrm{d} y}{\mathrm{~d} x}$$ is