Let the line $$\mathrm{L}$$ intersect the lines $$x-2=-y=z-1,2(x+1)=2(y-1)=z+1$$ and be parallel to the line $$\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$$. Then which of the following points lies on $$\mathrm{L}$$ ?

Let $$\overrightarrow{O A}=2 \vec{a}, \overrightarrow{O B}=6 \vec{a}+5 \vec{b}$$ and $$\overrightarrow{O C}=3 \vec{b}$$, where $$O$$ is the origin. If the area of the parallelogram with adjacent sides $$\overrightarrow{O A}$$ and $$\overrightarrow{O C}$$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $$O A B C$$ is equal to:

Let $$A=\{2,3,6,7\}$$ and $$B=\{4,5,6,8\}$$. Let $$R$$ be a relation defined on $$A \times B$$ by $$(a_1, b_1) R(a_2, b_2)$$ if and only if $$a_1+a_2=b_1+b_2$$. Then the number of elements in $$R$$ is __________.

Let $$\lim _\limits{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2 n}{\left(n^2+1\right) \sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+16}}\right.$$ $$\left.+\ldots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}}\right)$$ be $$\frac{\pi}{k}$$, using only the principal values of the inverse trigonometric functions. Then $$\mathrm{k}^2$$ is equal to _________.