Let $\mathrm{X}=\mathbf{R} \times \mathbf{R}$. Define a relation R on X as :

$$\left(a_1, b_1\right) R\left(a_2, b_2\right) \Leftrightarrow b_1=b_2$$

Statement I: $\quad \mathrm{R}$ is an equivalence relation.

Statement II : For some $(\mathrm{a}, \mathrm{b}) \in \mathrm{X}$, the $\operatorname{set} \mathrm{S}=\{(x, y) \in \mathrm{X}:(x, y) \mathrm{R}(\mathrm{a}, \mathrm{b})\}$ represents a line parallel to $y=x$.

In the light of the above statements, choose the correct answer from the options given below :

If the area of the region $\left\{(x, y):-1 \leq x \leq 1,0 \leq y \leq \mathrm{a}+\mathrm{e}^{|x|} \mid-\mathrm{e}^{-x}, \mathrm{a}>0\right\}$ is $\frac{\mathrm{e}^2+8 \mathrm{e}+1}{\mathrm{e}}$, then the value of $a$ is :

Let the range of the function $f(x)=6+16 \cos x \cdot \cos \left(\frac{\pi}{3}-x\right) \cdot \cos \left(\frac{\pi}{3}+x\right) \cdot \sin 3 x \cdot \cos 6 x, x \in \mathbf{R}$ be $[\alpha, \beta]$. Then the distance of the point $(\alpha, \beta)$ from the line $3 x+4 y+12=0$ is :

The system of equations

$$\begin{aligned} & x+y+z=6, \\ & x+2 y+5 z=9, \\ & x+5 y+\lambda z=\mu, \end{aligned}$$

has no solution if