Let the vectors $$\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}, \vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}$$ and $$\vec{c}=t \hat{i}-t \hat{j}+\hat{k}, t \in \mathbf{R}$$ be such that for $$\alpha, \beta, \gamma \in \mathbf{R}, \alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\overrightarrow{0} \Rightarrow \alpha=\beta=\gamma=0$$. Then, the set of all values of $$t$$ is :

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $$\cos ^{-1}(x)-2 \sin ^{-1}(x)=\cos ^{-1}(2 x)$$ is equal to :

Let the operations $$*, \odot \in\{\wedge, \vee\}$$. If $$(\mathrm{p} * \mathrm{q}) \odot(\mathrm{p}\, \odot \sim \mathrm{q})$$ is a tautology, then the ordered pair $$(*, \odot)$$ is :

Let a vector $$\vec{a}$$ has magnitude 9. Let a vector $$\vec{b}$$ be such that for every $$(x, y) \in \mathbf{R} \times \mathbf{R}-\{(0,0)\}$$, the vector $$(x \vec{a}+y \vec{b})$$ is perpendicular to the vector $$(6 y \vec{a}-18 x \vec{b})$$. Then the value of $$|\vec{a} \times \vec{b}|$$ is equal to :