Let two straight lines drawn from the origin $$\mathrm{O}$$ intersect the line $$3 x+4 y=12$$ at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ such that $$\triangle \mathrm{OPQ}$$ is an isosceles triangle and $$\angle \mathrm{POQ}=90^{\circ}$$. If $$l=\mathrm{OP}^2+\mathrm{PQ}^2+\mathrm{QO}^2$$, then the greatest integer less than or equal to $$l$$ is :

If the line $$\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z$$ makes a right angle with the line $$\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}$$, then $$4 \lambda+9 \mu$$ is equal to :

Consider the following two statements :

Statement I: For any two non-zero complex numbers $$z_1, z_2,(|z_1|+|z_2|)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right) \text {, and }$$

Statement II : If $$x, y, z$$ are three distinct complex numbers and $$\mathrm{a}, \mathrm{b}, \mathrm{c}$$ are three positive real numbers such that $$\frac{\mathrm{a}}{|y-z|}=\frac{\mathrm{b}}{|z-x|}=\frac{\mathrm{c}}{|x-y|}$$, then $$\frac{\mathrm{a}^2}{y-z}+\frac{\mathrm{b}^2}{z-x}+\frac{\mathrm{c}^2}{x-y}=1$$.

Between the above two statements,

Let A and B be two square matrices of order 3 such that $$\mathrm{|A|=3}$$ and $$\mathrm{|B|=2}$$. Then $$|\mathrm{A}^{\mathrm{T}} \mathrm{A}(\operatorname{adj}(2 \mathrm{~A}))^{-1}(\operatorname{adj}(4 \mathrm{~B}))(\operatorname{adj}(\mathrm{AB}))^{-1} \mathrm{AA}^{\mathrm{T}}|$$ is equal to :