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 :

The product of all solutions of the equation $\mathrm{e}^{5\left(\log _{\mathrm{e}} x\right)^2+3}=x^8, x>0$, is :