The equations of the sides AB, BC and CA of a triangle ABC are 2x +...

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

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

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

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

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

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

This point lies on the line $x + py = 39$

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

$⇒ p = 8$

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 = \dfrac{1}{2}\left|\begin{cases}1 & - 2 & 1 \\ 7 & 4 & 1 \\ \dfrac{- 13}{5} & \dfrac{26}{5} & 1\end{cases}\right|

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

$= 9\left[\dfrac{18}{5}\right] = \dfrac{162}{5} = 32 . 4 \notin \left(34, 38\right)$