Explanation:
(A) : The given equation is,
\(x^{2}-3 x y+\lambda y^{2}+3 x-5 y+2=0\)
Now, 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
\(\begin{aligned} & \left|\begin{array}{ccc}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right|=0 \\ & \left|\begin{array}{ccc}1 & -\frac{3}{2} & \frac{3}{2} \\ -\frac{3}{2} & \lambda & -\frac{5}{2} \\ \frac{3}{2} & -\frac{5}{2} & 2\end{array}\right|=0\end{aligned}\)
\(\therefore \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\left|\begin{array}{ccc}2 & -3 & 3 \\ -3 & 2 \lambda & -5 \\ 3 & -5 & 4\end{array}\right|=0\)
\(\therefore 2(8 \lambda-25)+3(-12+15)+3(15-6 \lambda)=0\)
\(16 \lambda-50+9+45-18 \lambda=0\)
\(\therefore-2 \lambda=-4 \quad \Rightarrow \lambda=2 \quad \Rightarrow \mathrm{b}-2\)
Now, \(\tan =\theta=\left|\frac{2 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right|\)
\(\left|\frac{2 \sqrt{\frac{9}{4}-2}}{1+2}\right|=\left|\frac{2 \sqrt{\frac{1}{4}}}{3}\right|\)
\(\tan \theta=\frac{1}{3} \Rightarrow \cot ^{2} \theta\)
Here \(\operatorname{cosec}^{2} \theta=1+\cot ^{2} \theta=1+9=10\)
