If for $\theta \in\left[-\frac{\pi}{3}, 0\right]$, the points $(x, y)=\left(3 \tan \left(\theta+\frac{\pi}{3}\right), 2 \tan \left(\theta+\frac{\pi}{6}\right)\right)$ lie on $x y+\alpha x+\beta y+\gamma=0$, then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Let the system of equations :

$$ \begin{aligned} & 2 x+3 y+5 z=9 \\ & 7 x+3 y-2 z=8 \\ & 12 x+3 y-(4+\lambda) z=16-\mu \end{aligned}$$

have infinitely many solutions. Then the radius of the circle centred at $(\lambda, \mu)$ and touching the line $4 x=3 y$ is :

Let the set of all values of $p \in \mathbb{R}$, for which both the roots of the equation $x^2-(p+2) x+(2 p+9)=0$ are negative real numbers, be the interval $(\alpha, \beta]$. Then $\beta-2 \alpha$ is equal to

Among the statements

(S1) : The set $\left\{z \in \mathbb{C}-\{-i\}:|z|=1\right.$ and $\frac{z-i}{z+i}$ is purely real $\}$ contains exactly two elements, and

(S2) : The set $\left\{z \in \mathbb{C}-\{-1\}:|z|=1\right.$ and $\frac{z-1}{z+1}$ is purely imaginary $\}$ contains infinitely many elements.