If direction cosines of two lines satisfy the equations $$l+m+n=0$$and $$amn+bnl+clm=0$$then show that the angle between them is $$\frac{\pi}{3}$$ if $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$$
My Attempt
Eliminating $l$ from the two equations I managed to obtain the quadratic equation $c\left(\frac{m}{n}\right)^2+(b+c-a)\left(\frac{m}{n}\right)+b=0$ If $l_{1},m_{1},n_{1}$ and $l_{2},m_{2},n_{2}$ be the two lines then $\frac{m_{1}}{n_{1}}$ and $\frac{m_{2}}{n_{2}}$ will be the two roots of the quadratic.Hence $\frac{m_{1}m_{2}}{n_{1}n_{2}}=\frac{b}{c}$ or $\frac{m_{1}m_{2}}{b}=\frac{n_{1}n_{2}}{c}$. Similarly by symmetry one can say that $\frac{l_{1}l_{2}}{a}=\frac{m_{1}m_{2}}{b}=\frac{n_{1}n_{2}}{c}=k$
Now $\cos \theta=\frac{1}{2}=l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}=k(a+b+c)$
So,$k=\frac{1}{2(a+b+c)}$
How to proceed from here