Three sides of a triangle have the equations $L_{r} = y - m_r x - C_{r} = 0; r = 1, 2, 3$. Then $\lambda L_{2}L_{3} + \mu L_{3}L_{1} + \gamma L_{1}L_{2} = 0$. where $\lambda \neq 0, \mu \neq 0, \gamma \neq 0$, is the equation of circumcircle of triangle if

Solution: Given that;

Equation of three sides of triangle: $L_{1}=y-m_{1} x-C_{1}=0$

$\begin{array}{l} L_{2}=y-m_{2} x-c_{2}=0 \\ L_{3}=y-m_{3} x-c_{3}=0 \end{array}$

Equation of circumcircle $=\lambda L_{2} L_{3}+\mu L_{3} L_{1}+\gamma L_{1} L_{2}=0$ Now,

since $\lambda L_{2} L_{3}+\mu L_{3} L+r L_{2} L_{1}=0$

$\begin{aligned} \Rightarrow & \lambda\left(y-m_{3} x-c_{3}\right)\left(y-m_{2} x-c_{2}\right)+\mu\left(y-m_{3} x-c_{3}\right)\left(y-m_{1} x-c_{1}\right)+ \\ & \gamma\left(y-m_{2} x-c_{2}\right)\left(y-m_{1} x-c_{1}\right)=0 \end{aligned}$

$\because$ This is the equation of a circle,

$\therefore$ Coefficient of $x^{2}=$ Coefficient of $y^{2} \&$ Coefficient of $x y=0$

$\begin{array}{l} \Rightarrow+\lambda m_{3} m_{2}+\mu\left(+m_{3} m_{1}\right)+\gamma\left(m_{1} m_{2}\right)=\lambda+\mu+\gamma \\ \Rightarrow \lambda\left(m_{3} m_{2}-1\right)+\mu\left(m_{3} m_{1}-1\right)+\gamma\left(m_{1} m_{2}-1\right)=0 \end{array}$

Also,

$\begin{array}{l} -\lambda m_{2}-\lambda m_{3}-\mu m_{3}-\mu m_{1}-\gamma m_{2}-\gamma m_{1}=0 \\ \Rightarrow \lambda m_{2}+\lambda m_{3}+\mu m_{3}+\mu m_{1}+\gamma m_{2}+\gamma m_{1}=0 \\ \Rightarrow \lambda\left(m_{2}+m_{3}\right)+\mu\left(m_{1}+m_{3}\right)+\gamma\left(m_{1}+m_{2}\right)=0 \end{array}$

Hence, option (A) is correct.