If the function $$f$$ is given by $$f(x)=x^3-3(a-2) x^2+3 a x+7$$, for some $$\mathrm{a} \in \mathbb{R}$$, is increasing in $$(0,1]$$ and decreasing in $$[1,5)$$, then a root of the equation $$\frac{\mathrm{f}(x)-14}{(x-1)^2}=0(x \neq 1)$$ is

The unit vector perpendicular to each of the vectors $$\bar{a}+\bar{b}$$ and $$\bar{a}-\bar{b}$$, where $$\bar{a}=\hat{i}+\hat{j}+\hat{k}$$ and $$\overline{\mathrm{b}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$$ is

The logical statement $$(\sim(\sim \mathrm{p} \vee \mathrm{q}) \vee(\mathrm{p} \wedge \mathrm{r})) \wedge(\sim \mathrm{q} \wedge \mathrm{r})$$ is equivalent to

Let $$\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \bar{b}=\hat{i}+\hat{j}$$ and $$\bar{c}$$ be a vector such that $$|\bar{c}-\bar{a}|=4,|(\bar{a} \times \bar{b}) \times \bar{c}|=3$$ and the angle between $$\overline{\mathrm{c}}$$ and $$\overline{\mathrm{a}} \times \overline{\mathrm{b}}$$ is $$\frac{\pi}{6}$$, then $$\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}$$ is equal to