Let $z_1, z_2$ and $z_3$ be three complex numbers on the circle $|z|=1$ with $\arg \left(z_1\right)=\frac{-\pi}{4}, \arg \left(z_2\right)=0$ and $\arg \left(z_3\right)=\frac{\pi}{4}$. If $\left|z_1 \bar{z}_2+z_2 \bar{z}_3+z_3 \bar{z}_1\right|^2=\alpha+\beta \sqrt{2}, \alpha, \beta \in Z$, then the value of $\alpha^2+\beta^2$ is :

Let $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\mathrm{L}_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $\mathrm{L}_1$ and $\mathrm{L}_2$ ?

Let $A=\{1,2,3, \ldots, 10\}$ and $B=\left\{\frac{m}{n}: m, n \in A, m< n\right.$ and $\left.\operatorname{gcd}(m, n)=1\right\}$. Then $n(B)$ is equal to :

Let the triangle PQR be the image of the triangle with vertices $(1,3),(3,1)$ and $(2,4)$ in the line $x+2 y=2$. If the centroid of $\triangle \mathrm{PQR}$ is the point $(\alpha, \beta)$, then $15(\alpha-\beta)$ is equal to :