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:

The line passing through the intersection points $P$ and $Q$ of two circles $S_1=0$ and $S_2=0$ is the radical axis: $S_1-S_2=0$.$(2ax-(-3ax))+(cy-dy)+(a-(-1))=0$$5ax+(c-d)y+a+1=0$ The given line is $5x+by-a=0$.For these two to be the same line:$\frac{5a}{5}=\frac{c-d}{b}=\frac{a+1}{-a}$ From $\frac{5a}{5}=\frac{a+1}{-a}⟹a=\frac{a+1}{-a}$$-a^2=a+1⟹a^2+a+1=0$.The discriminant of this quadratic in $a$ is $D=1^2-4(1)(1)=-3<0$.Since there are no real values for $a$, the condition is never satisfied.