The equations of the sides AB, BC and CA of a triangle ABC are $2\mathrm{x}+\mathrm{y}=0;\mathrm{x}+\mathrm{py}=15\mathrm{a}\text{ and }\mathrm{x}-\mathrm{y}=3$ respectively. If its orthocentre is $(2,\mathrm{a}),-\frac{1}{2}<\mathrm{a}<2$ then p is equal to

$$\begin{gathered} \text{ Coordinates of }\mathrm{A}(1,-2),\mathrm{B}\left(\frac{15\mathrm{a}}{1-2\mathrm{p}},\frac{-30\mathrm{a}}{1-2\mathrm{p}}\right)\text{ and } \\ \text{ orthocentre }\mathrm{H}(2,\mathrm{a}) \\ \text{ Slope of }\mathrm{AH}=\mathrm{p}, \boxed{\mathrm{a}+2=\mathrm{p}} \\ \text{ Slope of }\mathrm{BH}=-1 \\ 31\mathrm{a}-2\mathrm{ap}=15\mathrm{a}+4\mathrm{p}-2 \\ \Rightarrow 31\left(\mathrm{p}-2\right)-2\mathrm{p}\left(\mathrm{p}-2\right)=15\left(\mathrm{p}-2\right)+4\mathrm{p}-2 \\ \Rightarrow {\mathrm{p}}^2-8\mathrm{p}+15=0\therefore \mathrm{p}=3,5 \\ \Rightarrow \boxed{\mathrm{a}=1\mathrm{\&}\mathrm{p}=3} \end{gathered}$$