The point of intersection of the lines (a^3+3) x+a y+a-3=0 and...

Given equation of lines are $\left(a^3+3\right) x+a y+a-3=0 and \left(a^5+2\right) x+(a+2) y+2 a+3=0$ (a real) Since point of intersection of lines lies on $\mathrm{y}$-axis. $\therefore$ Put $x=0$ in each equation, we get $a \mathrm{y}+a-3=0 and (a+2) y+2 a+3=0$ On solving these we get $$\begin{aligned}& (a+2)(a-3)-a(2 a+3)=0 \\ & \Rightarrow a²-a-6-2 a²-3 a=0 \\ & \Rightarrow \quad-a²-4 a-6=0 \Rightarrow a²+4 a+6=0 \\ & \Rightarrow \quad a=\frac{-4 \pm \sqrt{16-24}}{2}=\frac{-4 \pm \sqrt{-8}}{2}\\ & (\text{ not real) } \end{aligned}$$This shows that the point of intersection of the lines lies on the $y$-axis for no value of ' $a$ '.