The circles $$x^2+y^2+2 \mathrm{a} x+\mathrm{c}=0$$ and $$x^2+y^2+2 b y+c=0$$ touch each other externally, if

Given $$\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2} & , \text { if } x<0 \\ \mathrm{a} & , \text { if } x=0 \\ \frac{\sqrt{x}}{\sqrt{16-\sqrt{x}-4}}, & \text { if } x>0\end{array}\right.$$

If $$\mathrm{f}(x)$$ is continuous at $$x=0$$, then value of a is

$$A, B, C, D$$ are four points in a plane with position vectors $$\bar{a}, \bar{b}, \bar{c}, \bar{d}$$ respectively such that $$(\bar{a}-\bar{d}) \cdot(\bar{b}-\bar{c})=(\bar{b}-\bar{d}) \cdot(\bar{c}-\bar{a})=0$$. The point $$D$$, then is the ___________ of $$\triangle \mathrm{ABC}$$

Two adjacent of sides parallelogram $$\mathrm{ABCD}$$ are given by $$\overline{\mathrm{AB}}=2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}$$ and $$\overline{A D}=-\hat{i}+2 \hat{j}+2 \hat{k}$$. The side $$A D$$ is rotated by angle $$\alpha$$ in plane of parallelogram so that $$\mathrm{AD}$$ becomes $$\mathrm{AD}^{\prime}$$. If $$\mathrm{AD}^{\prime}$$ makes a right angle with the side $$A B$$, then the cosine of the angle $$\alpha$$ is given by