The negation of inverse of $$\sim \mathrm{p} \rightarrow \mathrm{q}$$ is

Let $$\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$$ and $$\overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$$. If $$\overline{\mathrm{c}}$$ is a vector such that $$\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=|\overline{\mathrm{c}}|,|\overline{\mathrm{c}}-\overline{\mathrm{a}}|=2 \sqrt{2}$$ and the angle between $$\overline{\mathrm{a}} \times \overline{\mathrm{b}}$$ and $$\overline{\mathrm{c}}$$ is $$60^{\circ}$$. Then $$|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|=$$

$$\begin{aligned} & \text { If the function given by} \mathrm{f}(\mathrm{x}) \\ & =-2 \sin \mathrm{x} \quad-\pi \leq \mathrm{x}<-(\pi / 2) \\ & =a \sin x+b \quad-(\pi / 2)< x<(\pi / 2) \\ & =\cos x \quad(\pi / 2) \leq x \leq \pi \\ \end{aligned}$$

is continuous in $$[-\pi, \pi]$$, then the value of $$(3 a+2 b)^3$$ is

If $$\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$$, where $$x>0$$, then $$x=$$