The equations of the sides A B , B C and C A of a triangle A B C are 2 x + y = 0 ,   x + p y = 39 and x…

Intersection of $2x+y=0 \& x-y=3$ is $A\left(1,-2\right)$

Perpendicular bisector of $AB$ is $y-3=\frac{1}{2}\left(x-2\right)\Rightarrow x-2y+4=0$

Take image of $A$ along the perpendicular bisector of $AB$, we get 

$\frac{x-1}{1}=\frac{y+2}{-2}=\frac{-2\left(9\right)}{5}=-\frac{18}{5}$

i.e. $x=-\frac{13}{5}, y=\frac{26}{5}$

 $B\left(-\frac{13}{5},\frac{26}{5}\right)$

This point lies on the line $x+py=39$

i.e. $-\frac{13}{5}+p\left(\frac{26}{5}\right)=39\Rightarrow -1+2p=15$

$\Rightarrow p=8$

Solving $x+8y=39$ and $x-y=3$, we get $C\left(7,4\right)$

Now ${\left(AC\right)}^2={\left(1-7\right)}^2+{\left(-2-4\right)}^2=72=9p$

and $A{C}^2+p^2=72+64=136$

Also $\Delta ABC=\frac{1}{2}\left|\begin{array}{ccc}1& -2& 1\\ 7& 4& 1\\ \frac{-13}{5}& \frac{26}{5}& 1\end{array}\right|$

$=\frac{1}{2}\left[18+18\times \frac{13}{5}\right]$

$=9\left[\frac{18}{5}\right]=\frac{162}{5}=32.4\notin \left(34,38\right)$