The number of elements in the set $$S=\left\{x \in \mathbb{R}: 2 \cos \left(\frac{x^{2}+x}{6}\right)=4^{x}+4^{-x}\right\}$$ is :
Let $$\mathrm{A}(\alpha,-2), \mathrm{B}(\alpha, 6)$$ and $$\mathrm{C}\left(\frac{\alpha}{4},-2\right)$$ be vertices of a $$\triangle \mathrm{ABC}$$. If $$\left(5, \frac{\alpha}{4}\right)$$ is the circumcentre of $$\triangle \mathrm{ABC}$$, then which of the following is NOT correct about $$\triangle \mathrm{ABC}$$?
Let $$Q$$ be the foot of perpendicular drawn from the point $$P(1,2,3)$$ to the plane $$x+2 y+z=14$$. If $$R$$ is a point on the plane such that $$\angle P R Q=60^{\circ}$$, then the area of $$\triangle P Q R$$ is equal to :
If $$(2,3,9),(5,2,1),(1, \lambda, 8)$$ and $$(\lambda, 2,3)$$ are coplanar, then the product of all possible values of $$\lambda$$ is: