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

Given the equations of the sides $AB, BC$ and $CA$ of a triangle $ABC$ are $2x + y = 0, x + py = 15a$ and $x - y = 3$ respectively.

Now on solving equation $2x + y = 0, x + py = 15a$ and $x - y = 3$ we get coordinates of $A$\left(1, - 2\right), $B\left(\dfrac{15a}{1 - 2p}, \dfrac{- 30a}{1 - 2p}\right)$ and $C = \left(\dfrac{18p - 30}{p + 1}, \dfrac{15p - 33}{p + 1}\right)$

And let orthocentre be $H\left(2, a\right)$

Slope of $AH = \dfrac{a + 2}{1}$

Sloe of $BC = - \dfrac{1}{p}$

As $AH ⊥ BC$, so $p = a + 2$

Now coordinate of $C = \left(\dfrac{18p - 30}{p + 1}, \dfrac{15p - 33}{p + 1}\right)$

So, slope of $HC$

$= \dfrac{\dfrac{15p - 33}{p + 1} - a}{\dfrac{18p - 30}{p + 1} - 2} = \dfrac{15p - 33 - \left(p - 2\right)\left(p + 1\right)}{18p - 30 - 2p - 2}$

$= \dfrac{16p - {p}^{2} - 31}{16p - 32}$

Now $HC ⊥ AB$, so $\text{slope of}HC \times \text{slope of}AB\text{=-1}$

$⇒ \dfrac{16p - {p}^{2} - 31}{16p - 32} \times - 2 = - 1$

$⇒ {p}^{2} - 8p + 15 = 0$

$⇒ p = 3or5$

But if $p = 5$ then $a = 3$ not acceptable as $p = a + 2$

$∴ p = 3$