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. If its orthocentre is $$(2, a)$$, $$-\dfrac{1}{2} < a < 2$$, then $$p$$ is equal to ______.

Equation of side $$AB$$ is $$2x + y = 0 \; \Rightarrow \; y = -2x$$, so its slope is $$m_{AB} = -2$$.

Equation of side $$CA$$ is $$x - y = 3 \; \Rightarrow \; y = x - 3$$, so its slope is $$m_{CA} = 1$$.

Equation of side $$BC$$ is $$x + py = 15a$$ (slope will be obtained later).

Vertices are obtained by intersecting pairs of sides:
A is the intersection of $$AB$$ and $$CA$$.
  From $$y = -2x$$ and $$y = x - 3$$, we get $$-2x = x - 3 \;\Rightarrow\; 3x = 3 \;\Rightarrow\; x_A = 1, \; y_A = -2.$$

C is the intersection of $$CA$$ and $$BC$$.
  From $$x - y = 3 \;\Rightarrow\; x = 3 + y$$ and $$x + py = 15a$$:
  $$3 + y + py = 15a \;\Rightarrow\; y_C = \dfrac{15a - 3}{1 + p},$$
  $$x_C = 3 + y_C = 3 + \dfrac{15a - 3}{1 + p}.$$

B is the intersection of $$AB$$ and $$BC$$.
  Using $$y = -2x$$ in $$x + py = 15a$$:
  $$x - 2px = 15a \;\Rightarrow\; x_B = \dfrac{15a}{\,1 - 2p\,}, \quad y_B = -\,\dfrac{30a}{\,1 - 2p\,}.$$

The orthocentre is given as $$H(2,\,a)$$. Altitudes pass through $$H$$ and the corresponding vertex, and each altitude is perpendicular to the opposite side.

Slope of $$AB$$ is $$-2$$, hence the altitude through C has slope $$\dfrac12$$.
Therefore, $$\dfrac{a - y_C}{\,2 - x_C\,} = \dfrac12.$$

Substituting $$y_C, x_C$$:
$$\dfrac{a - \dfrac{15a - 3}{1 + p}}{\,2 - \Bigl(3 + \dfrac{15a - 3}{1 + p}\Bigr)} = \dfrac12.$$

Simplifying numerator and denominator gives
$$\dfrac{a(p - 14) + 3}{\,2 - p - 15a} = \dfrac12,$$
which reduces to
$$a(2p - 13) + p + 4 = 0 \quad -(1).$$

Slope of $$CA$$ is $$1$$, so the perpendicular slope is $$-1$$.
Thus $$\dfrac{a - y_B}{\,2 - x_B\,} = -1.$$

Substituting $$y_B, x_B$$:
$$\dfrac{a + \dfrac{30a}{1 - 2p}}{\,2 - \dfrac{15a}{1 - 2p}} = -1.$$

Simplifying gives
$$a(8 - p) - 2p + 1 = 0 \quad -(2).$$

From $$(2)$$, solve for $$a$$:
$$a = \dfrac{2p - 1}{8 - p}, \qquad p \neq 8.$$

Substitute this value of $$a$$ into $$(1)$$:

$$\bigl(2p - 1\bigr)(2p - 13) + (p + 4)(8 - p) = 0.$$

Expand and combine like terms:
$$(4p^2 - 28p + 13) + (-p^2 + 4p + 32) = 0,$$
$$3p^2 - 24p + 45 = 0,$$
$$p^2 - 8p + 15 = 0,$$
$$(p - 3)(p - 5) = 0.$$

Possible values are $$p = 3$$ or $$p = 5$$.

Corresponding $$a$$ values:
For $$p = 3$$: $$a = \dfrac{2(3) - 1}{8 - 3} = \dfrac{5}{5} = 1.$$

For $$p = 5$$: $$a = \dfrac{2(5) - 1}{8 - 5} = \dfrac{9}{3} = 3.$$

The given condition is $$-\dfrac12 \lt a \lt 2$$, which only $$a = 1$$ satisfies.
Hence $$p = 3$$ is the required value.

Final Answer: $$p = 3$$