Comparing given equation with $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$, we get
$\mathrm{a}=1, \mathrm{~h}=-\frac{-3}{2}, \mathrm{~b}=\lambda, \mathrm{g}=\frac{3}{2}, \mathrm{f}=\frac{-5}{2}, \mathrm{c}=2$
Given equation represents pair of lines only if $\left|\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right|=0$
$\begin{aligned}
&\left|\begin{array}{ccc}
1 & -\frac{3}{2} & \frac{3}{2} \\
3 & \lambda & -\frac{5}{2}
\end{array}\right|=0 \\
\therefore &\left|\begin{array}{ccc}
\frac{3}{2} & -\frac{5}{2} & 2
\end{array}\right| \\
\therefore & \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\left|\begin{array}{ccc}
2 & -3 & 3 \\
-3 & 2 \lambda & -5 \\
3 & -5 & 4
\end{array}\right|=0 \\
\therefore & 2(8 \lambda-25)+3(-12+15)+3(15-6 \lambda)=0 \\
& 16 \lambda-50+9+45-18 \lambda=0 \\
\therefore &-2 \lambda=-4 \quad \Rightarrow \lambda=2 \quad \Rightarrow \mathrm{b}=2
\end{aligned}$
$\text { Now, } \begin{aligned}
\tan \theta &=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right| \\
&=\left|\frac{2 \sqrt{\frac{9}{4}-2}}{1+2}\right|=\left|\frac{2 \sqrt{\frac{1}{4}}}{3}\right| \\
\tan \theta &=\frac{1}{3} \Rightarrow \cot \theta=3
\end{aligned}$
Here $\operatorname{cosec}^{2} \theta=1+\cot ^{2} \theta=1+9=10$