$\overline{\mathrm{a}} \mathrm{and} \overline{\mathrm{b}}$ are two vectors such that $\overline{\mathrm{a}} \mathrm{is} \mathrm{not} \mathrm{parallel} \mathrm{to} \overline{\mathrm{b}}$. If $\overline{\mathrm{p}}=(\mathrm{x}+2\mathrm{y}+3)\overline{\mathrm{a}}+(5\mathrm{x}-\mathrm{y}+2)\overline{\mathrm{b}}$ and $\overline{\mathrm{q}}=(2\mathrm{x}+3\mathrm{y}+5)\overline{\mathrm{a}}+(\mathrm{x}-5\mathrm{y}-2)\overline{\mathrm{b}}$ are two vectors such that $\overline{\mathrm{p}}=2\mathrm{q},\text{ then }\mathrm{x}-2\mathrm{y}=$

$\overline{\mathrm{p}}=(\mathrm{x}+2\mathrm{y}+3)\overline{\mathrm{a}}+(5\mathrm{x}-\mathrm{y}+2)\overline{\mathrm{b}}$ and $\overline{\mathrm{q}}=(2\mathrm{x}+3\mathrm{y}+5)\overline{\mathrm{a}}+(\mathrm{x}-5\mathrm{y}-2)\overline{\mathrm{b}}$ are two vectors then 

$\stackrel{-}{\mathrm{p}}=2\stackrel{-}{\mathrm{q}}$

$$\begin{gathered} x+2y+3=4x+3y+5 \\ 3x+y+2=0 \end{gathered}$$ and $$\begin{gathered} 5x-y+2=2x-10y-4 \\ 3x+9y+6=0 \\ x+3y+2=0 \end{gathered}$$

solving two equations will get,