Let $\overline{a},\overline{b}$ are two non collinear vectors if $\overline{OA}=\left(x+4y\right)\overline{a}+\left(2x+y+1\right)\overline{b}$, $\overline{OB}=\left(y-2x+2\right)\overline{a}+\left(2x-3y-1\right)\overline{b}$ and $3\overline{OA}=2\overline{OB}$ then $\left(x,y\right)=$   

Given $\overline{a},\overline{b}$ are two non collinear vectors. 

Given $\overline{OA}=\left(x+4y\right)\overline{a}+\left(2x+y+1\right)\overline{b}$ and $\overline{OB}=\left(y-2x+2\right)\overline{a}+\left(2x-3y-1\right)\overline{b}$

$3\overline{\mathrm{OA}}=2\overline{\mathrm{OB}}$

$(3x+12y)\overline{a}+(6x+3y+3)\overline{b}=(2y-4x+4)\overline{a}+(4x-6y-2)\overline{b}$

equating the coefficients of  $\overline{a},\overline{b}$ on both sides 

$7x+10y=4 ... \left(1\right)$

$2x+9y=5 ... \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$ we get $\boxed{\boxed{\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{1}}}$