If the circles $${x^2} + {y^2} + 2ax + cy + a = 0$$      and $${x^2} + {y^2} - 3ax + dy - 1 = 0$$      intersect in two distinct points $$P$$ and $$Q$$ then the line $$5x + by - a = 0$$    passes through $$P$$ and $$Q$$ for :

Solution :
$$\eqalign{ & {S_1} = {x^2} + {y^2} + 2ax + cy + a = 0 \cr & {S_2} = {x^2} + {y^2} - 3ax + dy - 1 = 0 \cr} $$
Equation of common chord of circles $${S_1}$$ and $${S_2}$$ is given by $${S_1} - {S_2} = 0$$
$$ \Rightarrow 5ax + \left( {c - d} \right)y + a + 1 = 0$$
Given that $$5x + by - a = 0$$    passes through $$P$$ and $$Q$$
$$\therefore $$  The two equations should represent the same line
$$\eqalign{ & \Rightarrow \frac{a}{1} = \frac{{c - d}}{b} = \frac{{a + 1}}{{ - a}} \cr & \Rightarrow a + 1 = - {a^2} \cr & \Rightarrow {a^2} + a + 1 = 0 \cr} $$
No real value of $$a.$$