If the three lines x-3 y=p, a x+2 y=q and a x+y=r form a...

Since three lines $x-3 y=p$, $a x+2 y=q$ and $a x+y=r$ form a right angled triangle $\therefore$ product of slopes of any two lines $=-1$ Suppose $a x+2 y=q$ and $x-3 y=p$ are $\perp$ to each other. \therefore $\frac{-a}{2} \times \frac{1}{3}=-$1 \Rightarrow a=6 Now, consider option one by one $a=6$ satisfies only option (a) $\therefore$ Required answer is $a^2-9 a+18=0$