1. Let the sides of triangle be:

$AB: 2x+y=0 \quad\Rightarrow\quad m_{AB}=-2$ $BC: x+py=15a \quad\Rightarrow\quad m_{BC}=-\frac{1}{p}$ $CA: x-y=3 \quad\Rightarrow\quad y=x-3 \Rightarrow m_{CA}=1$

The orthocentre is given as $H=(2,a)$.

2. Use the fact that the altitude from a vertex is perpendicular to the opposite side.

Since $CA$ has slope $1$, the altitude from $B$ has slope $-1$. Because this altitude passes through the orthocentre $H(2,a)$, its equation is

$y-a=-1(x-2)$ $x+y=a+2$

Now point $B$ is the intersection of sides $AB$ and $BC$, so it must lie on this altitude as well. Since $B$ lies on $AB: 2x+y=0$, intersect:

$2x+y=0$ $x+y=a+2$

Subtracting, we get

$x=-(a+2)$

Then

$y=2a+4$

So

$B\big(-(a+2),\,2a+4\big)$

3. Since $B$ also lies on $BC: x+py=15a$, substitute coordinates of $B$:

$-(a+2)+p(2a+4)=15a$ $-(a+2)+2p(a+2)=15a$ $(a+2)(2p-1)=15a$

So,

$2p-1=\frac{15a}{a+2} \quad \text{...(1)}$

4. Now use altitude from $A$.

Vertex $A$ is intersection of $AB$ and $CA$:

From $CA$, $y=x-3$ Substitute in $AB$:

$2x+(x-3)=0$ $3x=3 \Rightarrow x=1, \quad y=-2$

Thus,

$A=(1,-2)$

Altitude from $A$ passes through $H(2,a)$, so its slope is

$m_{AH}=\frac{a-(-2)}{2-1}=a+2$

Since this altitude is perpendicular to $BC$, whose slope is $-\frac{1}{p}$,

$m_{AH}\cdot m_{BC}=-1$ $(a+2)\left(-\frac{1}{p}\right)=-1$ $a+2=p$

So,

$p=a+2 \quad \text{...(2)}$

5. Substitute (2) into (1):

$(a+2)(2(a+2)-1)=15a$ $(a+2)(2a+3)=15a$ $2a^2+7a+6=15a$ $2a^2-8a+6=0$ $a^2-4a+3=0$ $(a-1)(a-3)=0$

So,

$a=1 \quad \text{or} \quad a=3$

Given,

$-\frac{1}{2}<a<2$

Hence,

$a=1$

Then from $p=a+2$,

$p=3$

6. Final answer:

$\boxed{3}$

7. Comparison with stored correct answer:

Stored correct answer = $3$. My derived answer also = $3$. So they agree.