If the three lines $x-3 y=p, a x+2 y=q$ and $a x+y=r$ form a right-angled triangle then :
If the three lines $x-3 y=p, a x+2 y=q$ and $a x+y=r$ form a right-angled triangle then :
$a^2-9 a+18=0$
$a^2-6 a-12=0$
$a^2-6 a-18=0$
$a^2-9 a+12=0$
Solution
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 $a^2-9 a+18=0$